datavorous/entrance-exam-dataset huggingface dataset:
The sum of the infinite series 1+23+732+1233+1734+2235+…… is equal to:

Options:
A 94
B 154
C 114
D 134

The correct answer is:
134

Solution:
S=1+23+732+1233+1734+…

S3=13+232+733+1234+…

2S3=1+13+532+533+534+…+ up to infinite terms

⇒2S3=43+5911-13

⇒2S3=136

⇒S=134

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2021 (26 Feb Shift 1)
The sum of the infinite series 13+33.7+53.7.11+73.7.11.15+....... is

Options:
A 12
B 13
C 16
D 14

The correct answer is:
12

Solution:
The nth term of the series is
tn=2n-13.7.11.---4n-1=124n-1-13.7.11.---4n-1
=1213.7.11.---4n-5-13.7.11.---4n-1
=12un-1-un, nâ¥2
Where, un=13.7.11.---4n-1
âSn=t1+âk=2ntk=13+12u1-u2+u2-u3---+un-1-un
=13+12u1-un=13+1213-un
âlimnââSn=13+16-12limnââun=12-0=12

Topic: Mathematics, Sequences and Series, JEE Main
The sum of the infinite series 1+56+1262+2263+3564+5165+7066+… is equal to:

Options:
A 425216
B 429216
C 288125
D 280125

The correct answer is:
288125

Solution:
Given,

S=1+56+1262+2263+3564+…. (i)

Multiplying with 16 both side

S6=16+562+1263+2264+… (ii)

On subtraction equation (i) - equation (ii)

We get, 56S=1+46+762+1063+1364+…(iii)

Again multiplying 16 both side,

536S=1+462+763+1064+1365+...(iv)

On subtraction equation (iii) - equation (iv)

2536S=1+36+362+363+…=85

S=288125

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2022 (29 Jun Shift 2)
The sum of the infinite series \( 1+\frac{2}{3}+\frac{7}{3^{2}}+\frac{12}{3^{3}}+\frac{17}{3^{4}}+\frac{22}{3^{5}}+\ldots \ldots \) is equal to:

Options:
A \( \frac{9}{4} \)
B \( \frac{15}{4} \)
C \( \frac{11}{4} \)
D \( \frac{13}{4} \)

The correct answer is:
\( \frac{13}{4} \)

Solution:
S=1+23+732+1233+1734+…

S3=13+232+733+1234+…

2S3=1+13+532+533+534+…+ up to infinite terms

⇒S=134

Topic: Mathematics, Sequences and Series, JEE Main
The sum of the intercepts of the plane on the coordinate axes, passing through the intersection of the planes 2x+3y+3z-5=0 and 2x-5y+3z+1=0 and parallel to the line x-12=y-2-5=z-3-7, is

Options:
A 25
B 11105
C 11102
D 3101

The correct answer is:
11102

Solution:
Equation of the plane passing through the line of intersection of the 2 given planes is 2x+3y+3z-5+Î»2x-5y+3z+1=0
âx2+2Î»+y3-5Î»+z3+3Î»+Î»-5=0
âµ This plane is parallel to x-12=y-2-5=z-3-7
â22+2Î»-53-5Î»-73+3Î»=0
â4+4Î»-15+25Î»-21-21Î»=0
â8Î»-32=0âÎ»=4
So, equation of the plane is 10x-17y+15z-1=0
Sum of intercepts =110-117+115=51-30+34510=55510=11102

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The sum of the intercepts on the coordinate axes made by a line passing through the point a,b and the common point of xa+yb=1 and xb+ya=1 is

Options:
A a-b2a+bab
B b-aa2-b2ab
C a+ba2-b2ab
D a-ba2+b2ab

The correct answer is:
b-aa2-b2ab

Solution:
Point of intersection of xa+yb=1 and xb+ya=1 is aba+b,aba+b
Now, line joining a,b and aba+b,aba+b is y-bx-a=aba+b-baba+b-a
ây-ba2=b2x-aâxab-ab+yba-ba=1
â sum of the x intercept and the y intercept is equal to ab-ab+ba-ba=b-aa2-b2ab

Topic: Mathematics, Straight Lines, JEE Main
The sum of the intercepts on the coordinate axes of the plane passing through the point –2,–2,2 and containing the line joining the points 1,–1,2 and 1,1,1 is

Options:
A 4
B 12
C -8
D -4

The correct answer is:
-4

Solution:
Equation of plane is
x+2y+2z-2-3-10-3-31=0
⇒-x+2+3y+2+6z-2=0
⇒x-3y-6z+8=0
∴ Sum of intercepts=-8+83+86=-4

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2018 (16 Apr Online)
The sum of the last eight coefficients in the expansion of $(1+x)^{15}$ is

Options:
A $2^{16}$
B $2^{15}$
C $2^{14}$
D None of these

The correct answer is:
$2^{14}$

Solution:
We have ${ }^{15} C_0+{ }^{15} C_1+\ldots+{ }^{15} C_{15}=2^{15}$
$\begin{aligned} & \Rightarrow 2\left({ }^{15} C_8+{ }^{15} C_9+\ldots+{ }^{15} C_{15}\right)=2^{15} \quad\left({ }^n C_r={ }^n C_{n-r}\right) \\
& \Rightarrow{ }^{15} C_8+{ }^{15} C_9+\ldots+{ }^{15} C_{15}=2^{14} \end{aligned}$

Topic: Mathematics, Binomial Theorem, JEE Main
The sum of the masses of reactants and products is equal in any physical or chemical reaction. This is in accordance with
law of:

Options:
A Multiple proportion
B Definite composition
C Conservation of mass
D Reciprocal proportion

The correct answer is:
Conservation of mass

Solution:
In every chemical reaction except nuclear reaction, Law of conservation of mass is followed i.e., mass can neither be created nor it can be destroyed but it can be transferred from one form to another form or during the chemical reaction, mass remains conserved.

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
The sum of the maximum and minimum values of the function fx=5x-7+x2+2x in the interval 54,2, where t is the greatest integer ≤t, is ______.

Options:

The correct answer is:
15

Solution:
Given,

fx=5x-7+x2+2x

=5x-7+x+12-1

=5x-7+x+12-1

as x-1=x-1 where . is greatest integer function

Now critical points of

fx=75,5-1,6-1,7-1,8-1,2

∴ Maximum or minimum value of fx occur at critical points or boundary points.

So, f54=34+4=194 and f75=0+4=4

As both 5x-7 and x2+2x are increasing in nature after x=75

So, f2=10-7+4+4=3+8=11

∴ f75min=4 and f2max=11

So sum of minimum and maximum value is 4+11=15

Topic: Mathematics, Functions, JEE Main, JEE Main 2022 (25 Jul Shift 2)
The sum of the minimum and maximum distance of the point $(4,-3)$ to the circle $x^2+y^2+4 x-10 y-7=0$, is

Options:
A $10$
B $12$
C $16$
D $20$

The correct answer is:
$20$

Solution:
Let the given point be $P(4,-3)$ and the given circle is
$x^2+y^2+4 x-10 y-7=0$
Centre of circle $=\mathrm{C}(-2,5)$
$\text { Radius }=\sqrt{(-2)^2+(5)^2+7}$
$=\sqrt{4+25+7}=\sqrt{36}=6$
Maximum distance, $a=C P+r$
Minimum distance, $b=C P-r$
Sum of the maximum and minimum distance,
$a+b=C P+r+C P-r=2 C P$
and $C P=\sqrt{(-2-4)^2+(5+3)^2}$
$=\sqrt{36+64}=\sqrt{100}=10$
Thus, $\quad a+b=2 C P=2 \times 10=20$

Topic: Mathematics, Circle, JEE Main
The sum of the number of σ and π bonds in 5-oxohexanoic acid is:

Options:
A 20
B 19
C 21
D 17

The correct answer is:
20

Solution:
(18σ, 2π)
5-oxo-hexanoic acid contains 18σ− and 2π− bonds. Hence, total number of σ− and π− bonds will be 20

Topic: Chemistry, General Organic Chemistry, JEE Main
The sum of the number of lone pairs of electrons on each central atom in the following species is \( \left[T e B r_{6}\right]^{2-},\left[B r F_{2}\right]^{+}, S N F_{3} \) and \( \left[X e F_{3}\right]^{-} \)[Atomic number : \( N=7, F=9, S=16, B r=35, T e=52, X e=54] \)

Options:

The correct answer is:
6

Solution:
Number of Ï-bonds formed by central metal atom/ ion	Number of lone pairs on central metal atom/ ion

(i) In â¡TeBr62-	6	1

(ii) Inâ¡BrF2+	2	2

(iii) Inâ¡SNF3	4	0

(iv) Inâ¡XeF3-	3	3

â Total number of lone pairs of electrons =1+2+0+3

=6

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The sum of the number of lone pairs of electrons on each central atom in the following species is

[TeBr6]2-,[BrF2]+ and [XeF3]-

(Atomic number: F = 9, Br = 35, Te = 52, Xe = 54)

Options:

The correct answer is:
6

Solution:
Species	Lone pair on central atom

TeBr2-	1

BrF2+	2

XeF3-	3

Total number of lone pair = 6.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The sum of the radii of inscribed and circumscribed circles for an $\mathrm{n}$ sided regular polygon of side a, is

Options:
A
$\frac{a}{4} \cot \left(\frac{\pi}{2 n}\right)$

B
$a \cot \left(\frac{\pi}{n}\right)$

C
$\frac{a}{2} \cot \left(\frac{\pi}{2 n}\right)$

D
$\operatorname{acot}\left(\frac{\pi}{2 n}\right)$

The correct answer is:
$\operatorname{acot}\left(\frac{\pi}{2 n}\right)$

Solution:
$\tan \left(\frac{\pi}{n}\right)=\frac{a}{2 r} ; \sin \left(\frac{\pi}{n}\right)=\frac{a}{2 R}$
$r+R=\frac{a}{2}\left[\cot \frac{\pi}{n}+\operatorname{cosec} \frac{\pi}{n}\right] \Rightarrow r+R=\frac{a}{2} \cdot \cot \left(\frac{\pi}{2 n}\right)$

Topic: Mathematics, Properties of Triangles, JEE Main, JEE Main 2003
The sum of the rational terms in the binomial expansion of $\left(2^{\frac{1}{2}}+3^{\frac{1}{5}}\right)^{10}$ is :

Options:
A
25

B
32

C
9

D
41

The correct answer is:
41

Solution:
$\left(2^{1 / 2}+3^{1 / 5}\right)^{10}={ }^{10} \mathrm{C}_0\left(2^{1 / 2}\right)^{10}$
$$
+{ }^{10} \mathrm{C}_1\left(2^{1 / 2}\right)^9\left(3^{1 / 5}\right)+\ldots \ldots+{ }^{10} \mathrm{C}_{10}\left(3^{1 / 5}\right)^{10}
$$
There are only two rational terms - first term
and last term.
Now sum of two rational terms
$$
=(2)^5+(3)^2=32+9=41
$$

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2013 (23 Apr Online)
The sum of the rational terms in the expansion of 2+31510 is equal to

Options:
A 31
B 41
C 51
D 61

The correct answer is:
41

Solution:
The general term is given by Tr+1=Cr10210−r23r5
This is rational if 10-r is even and r is a multiple of 5.
⇒r=0,10
So, the sum of the rational terms is
10C025+10C10×32=32+9=41

Topic: Mathematics, Binomial Theorem, JEE Main
The sum of the real roots of the equation x5-5x4+9x2-9x2+5x-1=0 is

Options:

The correct answer is:
4

Solution:
The sum of the coefficients is zero ⇒x=1 is a root ⇒x-1 is a factor.
x5-5x4+9x3-9x2+5x-1=x-1 x4-4x3+5x2-4x+1
Now, x4-4x3+5x2-4x+1=0
⇒x2-4x+5-4x+1x2=0
x+1x2-4x+1x+3=0
⇒x+1x=1,3
x+1x=1⇒x2-x+1=0⇒x=1±i32
x+1x=3⇒x2-3x+1=0⇒x=3+52
Roots are 1,1±i32,3±52
Sum of real roots =4

Topic: Mathematics, Quadratic Equation, JEE Main
The sum of the real roots of the equation
x-6-12-3xx-3-32xx+2=0, is equal to:

Options:
A 0
B -4
C 6
D 1

The correct answer is:
0

Solution:
Given x-6-12-3xx-3-32xx+2=0

On expansion, we get x-3xx+2-2xx-3+62x+2+3x-3-14x-9x=0

⇒x-3x2-6x-2x2+6x+62x+4+3x-9-1-5x=0

⇒-5x3+30x-30+5x=0

⇒5x3-35x+30=0

⇒x3-7x+6=0

⇒x-1x-2x+3

⇒x=1 or x=2 or x=-3.

Here all roots 1, 2, -3 are real, hence the sum of the real roots is 1+2-3=0.

Topic: Mathematics, Determinants, JEE Main, JEE Main 2019 (10 Apr Shift 2)
The sum of the real values of $x$ for which the middle term in the binomial expansion of $\left(\frac{x^{3}}{3}+\frac{3}{x}\right)^{8}$ equals 5670 is :

Options:
A 0
B 6
C 4
D 8

The correct answer is:
0

Solution:
Middle Term, $\left(\frac{n}{2}+1\right)^{\text {th }}$ term in the binomial

expansion of $\left(\frac{x^{3}}{3}+\frac{3}{x}\right)^{8}$ is,

$T_{4+1}={ }^{8} C_{4}\left(\frac{x^{3}}{3}\right)^{4}\left(\frac{3}{x}\right)^{4}=5670$

$\Rightarrow \quad \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2} \times x^{12-4}=5670$

$\Rightarrow \quad x^{8}=81$

$\Rightarrow x^{8}-81=0$

$\therefore$ sum of all values of $x=$ sum of roots of equation $\left(x^{8}-81=\right.$ 0).

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2019 (11 Jan Shift 1)
The sum of the reciprocals of focal distances of a focal chord $P Q$ of $y^{2}=4 a x$ is

Options:
A $\frac{1}{a}$
B $a$
C $2 a$
D $\frac{1}{2 a}$

The correct answer is:
$\frac{1}{a}$

Solution:
Equation of parabola $y^{2}=4 a x$
Total length of focal chord $=4 a$
and half length of focal chord $=2 a$
Now, sum of the reciprocals of focal distances
$$
=\frac{1}{2 a}+\frac{1}{2 a}=\frac{2}{2 a}=\frac{1}{a}
$$

Topic: Mathematics, Parabola, JEE Main
The sum of the roots of the equation 233x-2+211x+2=222x+1+1 is

Options:
A 111
B 211
C 311
D 411

The correct answer is:
211

Solution:
Let, t=211x
⇒211x322+211x.22=211x2.2+1
⇒t34+4t=2t2+1
⇒t3-8t2+16t-4=0
Cubic in t has roots t1,t2,t3
i.e. t1t2t3=4⇒211x1.211x2.211x3=4
⇒211x1+x2+x3=22
⇒11x1+x2+x3=2⇒x1+x2+x3=211

Topic: Mathematics, Quadratic Equation, JEE Main
The sum of the roots of the equation 3cos⁡x-sin⁡x=2 in 0,4π is kπ, then the value of 6k is

Options:

The correct answer is:
56

Solution:
3cos⁡x-sin⁡x=±2
32cos⁡x-12sin⁡x=±1
cosx+π6=±1
⇒x+π6=nπ
⇒x=nπ-π6
⇒x=π-π6,2π-π6,3π-π6,4π-π6
⇒ The sum of roots is 10π-2π3=283π
⇒6k=6×283=56

Topic: Mathematics, Trigonometric Equations, JEE Main
The sum of the roots of the equation cos⁡4x+6=7cos⁡2x in the interval 0,314 is λπ, then the numerical value of λ is

Options:
A 4950
B 2475
C 9900
D 4945

The correct answer is:
4950

Solution:
2cos2â¡2x-1+6=7cosâ¡2x
On putting cosâ¡2x=t, we get,
2t2-1+6=7t
2t2-7t+5=0
2t-5t-1=0
t=52,1
t=52 (not possible)
t=1âcos2x=1â2x=2nÏ
âx=nÏ
The roots in 0,314 are
Ï,2Ï,3Ï,.....,99Ï 100Ï>314
Sum of roots =Ï+2Ï+3Ï+....+99Ï=4950Ï
âÎ»=4950

Topic: Mathematics, Trigonometric Equations, JEE Main
The sum of the roots of the equation, x+1-2log23+2x+2log410-2-x=0, is :

Options:
A log214
B log212
C log213
D log211

The correct answer is:
log211

Solution:
Given, x+1-2log23+2x+2log410-2-x=0

⇒x+1-2log23+2x+2log2210.2x-12x=0

⇒x+1-2log23+2x+log210.2x-12x=0

⇒x+1-2log23+2x+log210.2x-1-log22x=0

⇒x+1-log23+2x2+log210.2x-1-xlog22=0

⇒x+1-log23+2x2+log210.2x-1-x=0

⇒x+1+log210.2x-13+2x2-x=0

⇒1+log210.2x-13+2x2=0

⇒log210.2x-13+2x2=-1

⇒log210.2x-19+2x2+6.2x=log212

⇒10.2x-19+2x2+6.2x=12

⇒210.2x-1=9+2x2+6.2x

⇒2x2-14.2x+11=0

Let 2x=y

⇒y2-14y+11=0

Let roots are y1=2x1 & y2=2x2

Product of roots, y1y2=2x1×2x2=2x1+x2=11

⇒log2x1+x2=log11

⇒x1+x2log2=log11

⇒x1+x2=log11log2

Hence, x1+x2=log211

Topic: Mathematics, Basic of Mathematics, JEE Main, JEE Main 2021 (31 Aug Shift 2)
The sum of the roots of the equation, $\mathrm{x}^2+|2 \mathrm{x}-3|-4=0$, is:

Options:
A
2

B
$-2$

C
$\sqrt{2}$

D
$-\sqrt{2}$

The correct answer is:
$\sqrt{2}$

Solution:
$$
\begin{aligned}
&x^2+|2 x-3|-4=0 \\
&|2 x-3|=\left\{\begin{array}{lll}
(2 x-3) & \text { if } & \mathrm{x}>\frac{3}{2} \\
-(2 x-3) & \text { if } & \mathrm{x} < \frac{3}{2}
\end{array}\right.
\end{aligned}
$$
$$
\begin{aligned}
&\text { for } x>\frac{3}{2}, \quad x^2+2 x-3-4=0 \\
&x^2+2 x-7=0 \\
&x=\frac{-2 \pm \sqrt{4+28}}{2} \\
&=\frac{-2 \pm 4 \sqrt{2}}{2}=-1 \pm 2 \sqrt{2} \\
&\text { Here } x=2 \sqrt{2}-1 \\
&\left\{2 \sqrt{2}-1 < \frac{3}{2}\right\} \\
&\text { for } x < \frac{3}{2} \\
&x^2-2 x+3-4=0 \\
&\Rightarrow x^2-2 x-1=0 \\
&\Rightarrow x=\frac{2 \pm \sqrt{4+4}}{2}=\frac{2 \pm 2 \sqrt{2}}{2}=1 \pm \sqrt{2} \\
&\text { Here } x=1-\sqrt{2} \quad\left\{(1-\sqrt{2}) < \frac{3}{2}\right\}
\end{aligned}
$$
Sum of roots: $(2 \sqrt{2}-1)+(1-\sqrt{2})=\sqrt{2}$

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2014 (12 Apr Online)
The sum of the roots of the equation x2-x-6=x+2 is

Options:
A 0
B -2
C 2
D 4

The correct answer is:
4

Solution:
x2-x-6=x+2x-3
Case i x<-2 or x≥3
x2-x-6=x+2⇒x2-2x-8=0
x-4x+2=0⇒x=-2,4
⇒x=4 in this case
Case ii-2≤x<3
-x2+x+6=x+2⇒x2=4⇒x=±2
⇒x=-2,2 in this case
∴ The roots are -2,2,4
⇒sum of roots=4

Topic: Mathematics, Quadratic Equation, JEE Main
The sum of the series:
$1+\frac{1}{1+2}+\frac{1}{1+2+3}+\ldots \ldots$. upto 10 terms, is:

Options:
A
$\frac{18}{11}$

B
$\frac{22}{13}$

C
$\frac{20}{11}$

D
$\frac{16}{9}$

The correct answer is:
$\frac{20}{11}$

Solution:
$\mathrm{T}_r=\frac{1}{1+2+3+\ldots+r}=\frac{2}{r(r+1)}$
$\mathrm{S}_{10}=2 \sum_{r=1}^{10} \frac{1}{r(r+1)}=2 \sum_{r=1}^{10}\left[\frac{r+1}{r(r+1)}-\frac{r}{r(r+1)}\right]$
$=2 \sum_{r=1}^{10}\left(\frac{1}{r}-\frac{1}{r+1}\right)$
$=2\left[\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\ldots+\left(\frac{1}{10}-\frac{1}{11}\right)\right]$
$=2\left[1-\frac{1}{11}\right]=2 \times \frac{10}{11}=\frac{20}{11}$

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2013 (09 Apr Online)
The sum of the series 11−3⋅12+14+21−3⋅22+24+31−3⋅32+34+.... up to 10 terms is

Options:
A 45109
B -45109
C 55109
D -55109

The correct answer is:
-55109

Solution:
General term of the given series is given by,

Tn=n1-3n2+n4

⇒Tn=nn4-2n2+1-n2

⇒Tn=nn2-12-n2

⇒Tn=nn2-1-nn2-1+n

⇒Tn=12n2-1+n-n2-1-nn2-1-nn2-1+n

⇒Tn=121n2-1-n-1n2-1+n

⇒∑n=110Tn=121-1-11+11-15+15-111+...+189-1109

⇒∑n=110Tn=12-1-1109

⇒∑n=110Tn=-55109

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2024 (31 Jan Shift 1)
The sum of the series
$$
\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\ldots
$$
upto 15 terms is

Options:
A
1

B
2

C
3

D
4

The correct answer is:
3

Solution:
Given series is
$$
\begin{aligned}
& \frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\ldots . . \\
& n^{\text {th }} \text { term }=\frac{1}{\sqrt{n}+\sqrt{n+1}} \\
& \therefore 15^{\text {th }} \text { term }=\frac{1}{\sqrt{15}+\sqrt{16}}
\end{aligned}
$$
Thus, given series upto 15 terms is
$$
\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\ldots \ldots+\frac{1}{\sqrt{15}+\sqrt{16}}
$$
This can be re-written as
$$
\frac{1-\sqrt{2}}{-1}+\frac{\sqrt{2}-\sqrt{3}}{-1}+\frac{\sqrt{3}-\sqrt{4}}{-1}+\ldots \ldots+\frac{\sqrt{15}-\sqrt{16}}{-1}
$$
(By rationalization)
$$
\begin{gathered}
=-1+\sqrt{2}-\sqrt{2}+\sqrt{3}-\sqrt{3}+\sqrt{4}+\ldots . .-\sqrt{14}+\sqrt{15} \\
-\sqrt{15}+\sqrt{16} \\
=-1+\sqrt{16}=-1+4=3
\end{gathered}
$$
Hence, the required sum $=3$

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2012 (12 May Online)
The sum of the series $1^2+2.2^2+3^2+2.4^2+5^2+2.6^2+\ldots . .+2(2 m)^2$ is

Options:
A
$m(2 m+1)^2$

B
$m^2(m+2)$

C
$m^2(2 m+1)$

D
$m(m+2)^2$

The correct answer is:
$m(2 m+1)^2$

Solution:
The sum of the given series $1^2+2.2^2+3^2+$ $2.4^2+5^2+2.6^2+\ldots \ldots . .+2(2 m)^2$ is $\frac{2 m\left(2 m+1^2\right.}{2} \leqslant m\left(2 m+1^2\right)$

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2012 (07 May Online)
The sum of the series 1232+2132+32+3232+...... to 10 terms is

Options:
A 13909
B 17909
C 19909
D 22909

The correct answer is:
22909

Solution:
1232+2132+32+3232+........ to 10 terms
=532+732+932+1132........ to 10 terms
=52+72+92+112+.....9 to 10 terms
=19∑r=1102r+32
=194∑r=110r2+12∑r=110r+9∑r=1101
=1924×10×11×21763+612×10×112+9×10
=191540+660+90=22909

Topic: Mathematics, Sequences and Series, JEE Main
The sum of the series $1+\frac{2}{3}\left(\frac{1}{8}\right)+\frac{2 \times 5}{3 \times 6}\left(\frac{1}{8}\right)^2+\frac{2 \times 5 \times 8}{3 \times 6 \times 9}\left(\frac{1}{8}\right)^3+\ldots$ is

Options:
A $\frac{4}{\sqrt[3]{49}}$
B $\frac{\sqrt[3]{49}}{4}$
C $\frac{4}{\sqrt[3]{81}}$
D $\frac{\sqrt[3]{81}}{4}$

The correct answer is:
$\frac{4}{\sqrt[3]{49}}$

Solution:
$$
\begin{aligned}
& \text { Let } S=1+\frac{2}{3}\left(\frac{1}{8}\right)+\frac{2 \times 5}{3 \times 6}\left(\frac{1}{8}\right)^2+\frac{2 \times 5 \times 8}{3 \times 6 \times 9}\left(\frac{1}{8}\right)^3+\ldots \\
& =1+\frac{\frac{2}{3}}{1}\left(\frac{1}{8}\right)+\frac{\left(\frac{2}{3}\right)\left(\frac{5}{3}\right)}{2 !}\left(\frac{1}{8}\right)^2+\frac{\left(\frac{2}{3}\right)\left(\frac{5}{3}\right)\left(\frac{8}{3}\right)}{3 !}\left(\frac{1}{8}\right)^3+\ldots \\
& =\left(1-\frac{1}{8}\right)^{-\frac{2}{3}}\left[\because(1-x)^{-n}=1+n x+\frac{n(n+1)}{2 !} x^2+\ldots\right] \\
& =\left(\frac{7}{8}\right)^{-\frac{2}{3}}=\left(\frac{8}{7}\right)^{\frac{2}{3}}=\left(\frac{64}{49}\right)^{\frac{1}{3}}=\frac{(64)^{\frac{1}{3}}}{\sqrt[3]{49}}=\frac{4}{\sqrt[3]{49}}
\end{aligned}
$$

Topic: Mathematics, Binomial Theorem, JEE Main
The sum of the series $1+\frac{4}{3}+\frac{10}{9}+\frac{28}{27}+\ldots$ upto $n$ terms is

Options:
A
$\frac{7}{6} n+\frac{1}{6}-\frac{2}{3.2^{n-1}}$

B
$\frac{5}{3} n-\frac{7}{6}+\frac{1}{2.3^{n-1}}$

C
$n+\frac{1}{2}-\frac{1}{2.3^n}$

D
$n-\frac{1}{3}-\frac{1}{3.2^{n-1}}$

The correct answer is:
$n+\frac{1}{2}-\frac{1}{2.3^n}$

Solution:
Given series is
$1+\frac{4}{3}+\frac{10}{9}+\frac{28}{27}+\ldots . n$ terms
$=1+\left(1+\frac{1}{3}\right)+\left(1+\frac{1}{9}\right)+\left(1+\frac{1}{27}\right)+\ldots . n$ terms $=(1+1+1+\ldots+n$ terms $)$ $+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\ldots n\right.$ terms $)$ $=n+\frac{\frac{1}{3}\left(1-\frac{1}{3^n}\right)}{1-\frac{1}{3}}=n+\frac{1}{3} \times \frac{3}{2}\left[1-3^{-n}\right]$ $=n+\frac{1}{2}\left[1-3^{-n}\right]=n+\frac{1}{2}-\frac{1}{2.3^n}$

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2012 (19 May Online)
The sum of the series $1+\frac{1}{4.2 !}+\frac{1}{16.4 !}+\frac{1}{64.6 !}+\ldots \ldots \ldots$ ad inf. is

Options:
A
$\frac{\mathrm{e}-1}{\sqrt{\mathrm{e}}}$

B
$\frac{\mathrm{e}+1}{\sqrt{\mathrm{e}}}$

C
$\frac{\mathrm{e}-1}{2 \sqrt{\mathrm{e}}}$

D
$\frac{e+1}{2 \sqrt{e}}$

The correct answer is:
$\frac{e+1}{2 \sqrt{e}}$

Solution:
$\frac{e^x+e^{-x}}{2}=1+\frac{x^2}{2 !}+\frac{x^4}{4 !}+\frac{x^6}{6 !}+\ldots \ldots$
putting $x=1 / 2$ we get
$\frac{e+1}{2 \sqrt{e}}$

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2005
The sum of the series 1x+1+2x2+1+22x4+1+…+2100x2100+1 when x=2 is:

Options:
A 1-21014101-1
B 1+21014101-1
C 1+21004101-1
D 1-21004201-1

The correct answer is:
1-21014101-1

Solution:
Adding and subtracting 1x-1

S=1x-1-1x-1+1x+1+2x2+1+22x4+1+…+2100x2100+1

S=1x-1+1x+1-1x-1+2x2+1+22x4+1+…+2100x2100+1

S=1x-1+2x2+1+-2x2-1+22x4+1+….+2100x2100+1

S=1x-1+-22x4-1+22x4+1+….+2100x2100+1

S=1x-1-2101x2101-1

For x=2⇒S=1-210122101-1=1-21014101-1

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2021 (26 Aug Shift 1)
The sum of the series 21Â·2+ 52Â·3Â·2 + 103Â·4Â·22 + 174Â·5Â·23 + ...... to n terms is

Options:
A nn+1·2n+1
B n+1n·2n+1
C nn+1·2n
D n+1n·2n

The correct answer is:
nn+1·2n

Solution:
The required sum
=âr=1nr2+1rr+12r-1=âr=1n2rr+1-r-1r2r-1
= âr=1nrr+12r-r-1r2r-1= nn+1Â·2n

Topic: Mathematics, Sequences and Series, JEE Main
The sum of the series $\frac{1}{2 !}-\frac{1}{3 !}+\frac{1}{4 !}-\ldots$ upto infinity is

Options:
A
$e^{-2}$

B
$e^{-1}$

C
$\mathrm{e}^{-1 / 2}$

D
$\mathrm{e}^{1 / 2}$

The correct answer is:
$e^{-1}$

Solution:
$e^{-x}=1-x+\frac{x^2}{2 !}-\frac{x^3}{3 !}+\frac{x^4}{4 !}-\cdots$
put $x=1$
$\frac{1}{2 !}-\frac{1}{3 !}+\frac{1}{4 !} \cdots=e^{-1}$.

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2007
The sum of the series :
$(2)^2+2(4)^2+3(6)^2+\ldots$ upto 10 terms is :

Options:
A
11300

B
11200

C
12100

D
12300

The correct answer is:
12100

Solution:
$2^2+2(4)^2+3(6)^2+\ldots \ldots$ upto 10 terms $=2^2\left[1^3+2^3+3^3+\ldots \ldots\right.$ upto 10 terms $]$
$=4 \cdot\left(\frac{10 \times 11}{2}\right)^2=12100$

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2013 (23 Apr Online)
The sum of the series 2.20C0+5.20C1+8.20C2+11.20C3+.......+62.20C20 is equal to

Options:
A 226
B 225
C 224
D 223

The correct answer is:
225

Solution:
Let,

S= 2.20C0+5.20C1+8.20C3+.......+62.20C20 ...i
S=62.20C0+59.20C1+56.20C3+.......+2.20C20 ...ii (Writing in reverse order nCr=nCn-r )
Adding i and ii
2S=64(20C0+20C1+20C2+.....+20C20)
⇒S=32.220=225

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2019 (08 Apr Shift 1)
The sum of the series 3+8+16+27+41....... upto 20 terms is equal to

Options:
A 4230
B 4430
C 4330
D 4500

The correct answer is:
4430

Solution:
Sn=3+8+16+27+........+tn
Sn= 3+8+16+........+tn-1+tn

Subtracting, we get,
0=3+5+8+11+........+tn-tn-1-tn
⇒tn=1+2+5+8+11+.........n terms
=1+n24+n-13=1+n23n+1
Sn=Σtn=Σ1+32Σn2+12Σn
=32nn+12n+16+12nn+12+n
for n=20, S=20×21×414+20×214+20
=20×21441+1+20=4430

Topic: Mathematics, Sequences and Series, JEE Main
The sum of the series 512⋅42+1142⋅72+1772⋅102+.……. is

Options:
A 12
B 13
C 2
D 3

The correct answer is:
13

Solution:
S=512⋅42+1142⋅72+1772⋅102+.…3S=3.512⋅42+3⋅1142⋅72+3⋅1772⋅102+.….
3S=4-14+112⋅42+7-47+442·72+10-710+772⋅102+.….
3S=42-1212⋅42+72-4242⋅72+102-7272⋅102+.….
=112-142+142-172+172-1102+.….
⇒3S=1
S=13

Topic: Mathematics, Sequences and Series, JEE Main
The sum of the series $\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{3.4} \ldots \ldots \ldots \ldots \ldots \ldots .$. up to $\infty$ is equal to

Options:
A
$\log _e\left(\frac{4}{e}\right)$

B
$2 \log _e 2$

C
$\log _{\mathrm{e}} 2-1$

D
$\log _{\mathrm{e}} 2$

The correct answer is:
$\log _e\left(\frac{4}{e}\right)$

Solution:
$\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{3.4} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots\infty$
Let $T_n=\frac{1}{n(n+1)}=\left(\frac{1}{n}-\frac{1}{n+1}\right)$
$\mathrm{S}=\mathrm{T}_1-\mathrm{T}_2+\mathrm{T}_3-\mathrm{T}_4+\mathrm{T}_5 \ldots \ldots \ldots \ldots \ldots$
$=\left(\frac{1}{1}-\frac{1}{2}\right)-\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)-\left(\frac{1}{4}-\frac{1}{5}\right) \ldots \ldots \ldots$.
$=1-2\left[\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5} \ldots \ldots \ldots \ldots \infty\right]$
$=1-2[-\log (1+1)+1]=2 \log 2-1=\log \left(\frac{4}{e}\right)$

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2003