datavorous/entrance-exam-dataset huggingface dataset:
The sum of the series
${ }^{20} \mathrm{C}_0-{ }^{20} \mathrm{C}_1+{ }^{20} \mathrm{C}_2-{ }^{20} \mathrm{C}_3+\ldots-\ldots+{ }^{20} \mathrm{C}_{10} \text { is }$

Options:
A
$-{ }^{20} \mathrm{C}_{10}$

B
$\frac{1}{2}{ }^{20} \mathrm{C}_{10}$

C
$0$

D
${ }^{20} \mathrm{C}_{10}$

The correct answer is:
$\frac{1}{2}{ }^{20} \mathrm{C}_{10}$

Solution:
$(1+\mathrm{x})^{20}={ }^{20} \mathrm{C}_0+{ }^{20} \mathrm{C}_1 \mathrm{x}+\ldots+{ }^{20} \mathrm{C}_{10} \mathrm{x}{ }^{10}+\ldots+{ }^{20} \mathrm{C}_{20} \mathrm{x}^{20}$
put $\mathrm{x}=-1$,
$0={ }^{20} \mathrm{C}_0-{ }^{20} \mathrm{C}_1+\ldots-{ }^{20} \mathrm{C}_9+{ }^{20} \mathrm{C}_{10}-{ }^{20} \mathrm{C}_{11}+\ldots+{ }^{20} \mathrm{C}_{20}$
$0=2\left({ }^{20} \mathrm{C}_0-{ }^{20} \mathrm{C}_1+\ldots-{ }^{20} \mathrm{C}_9\right)+{ }^{20} \mathrm{C}_{10}$
$\Rightarrow{ }^{20} \mathrm{C}_0-{ }^{20} \mathrm{C}_1+\ldots+{ }^{20} \mathrm{C}_{10}=\frac{1}{2}{ }^{20} \mathrm{C}_{10}$.

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2007
The sum of the series $3.6+4.7+5.8+\ldots \ldots \ldots$. uppto $(n-2)$ terms

Options:
A $n^3+n^2+n+2$
B $\frac{1}{6}\left(2 n^3+12 n^2+10 n-84\right)$
C $n^3+n^2+n$
D None of these

The correct answer is:
$\frac{1}{6}\left(2 n^3+12 n^2+10 n-84\right)$

Solution:
$S=3.6+4.7+\ldots \ldots .$. upto $n-2$ terms
$=\left(1 \cdot 4+2 \cdot 5+3 \cdot 6+4.7+\ldots \ldots \ldots\right.$ upto $^n$ terms $)-14$
$=\Sigma n(n+3)-14=\frac{1}{6}\left(2 n^3+12 n^2+10 n\right)-14$
$=\left(\frac{2 n^3+12 n^2+10 n-84}{6}\right){ }_{\text {where }} n=3,4,5 \ldots \ldots$.
Trick : $S_1=18, S_2=46$
Now put in options $(n-2)=1,2$ i.e. $n=3,4$
Obviously (2) gives the values.

Topic: Mathematics, Sequences and Series, JEE Main
The sum of the series $1.2 .3+2.3 .4+3.4 .5+\ldots \ldots .$. to $n$ terms is

Options:
A $n(n+1)(n+2)$
B $(n+1)(n+2)(n+3)$
C $\frac{1}{4} n(n+1)(n+2)(n+3)$
D $\frac{1}{4}(n+1)(n+2)(n+3)$

The correct answer is:
$\frac{1}{4} n(n+1)(n+2)(n+3)$

Solution:
$T_n=n(n+1)(n+2)=n\left(n^2+3 n+2\right)=n^3+3 n^2+2 n$
$\therefore S_n=\Sigma\left(n^3\right)+\Sigma\left(3 n^2\right)+\Sigma(2 n)$
$S_n=\left[\frac{n(n+1)}{2}\right]^2+\frac{3 \cdot n(n+1)(2 n+1)}{6}+\frac{2 \cdot n(n+1)}{2}$
$S_n=\frac{1}{4} n(n+1)(n+2)(n+3)$

Topic: Mathematics, Sequences and Series, JEE Main
The sum of the series C020-C120+C220-C320+…+C1020 is equal to

Options:
A -C1020
B 12C1020
C 0
D C1020

The correct answer is:
12C1020

Solution:
S=C020-C120+C220+…+C1020
We know that,
C020-C120+C220-…+C20=020
⇒2C020-C1+20C2+20…-C920+C10=020
⇒C020-C120+C220…-C920=-12C1020
So, S=-12C1020+C1020=12C1020

Topic: Mathematics, Binomial Theorem, JEE Main
The sum of the series
$\sum_{r=0}^{n}(-1)^{r n} C_{r}\left(\frac{1}{2^{r}}+\frac{3^{r}}{2^{2 r}}+\frac{7^{r}}{2^{3 r}}+\frac{15^{r}}{2^{4 r}}+\ldots m\right.$ terms $)$ is

Options:
A $\frac{2^{m n}-1}{2^{m n}\left(2^{n}-1\right)}$
B $\frac{2^{m n}-1}{2^{n}-1}$
C $\frac{2^{m n}+1}{2^{n}+1}$
D None of these

The correct answer is:
$\frac{2^{m n}-1}{2^{m n}\left(2^{n}-1\right)}$

Solution:
$\quad \sum_{\mathrm{r}=0}^{\mathrm{n}}(-1)^{\mathrm{r}} \cdot{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}$
$\left(\frac{1}{2^{\mathrm{r}}}+\frac{3^{\mathrm{r}}}{2^{2 \mathrm{r}}}+\frac{7^{\mathrm{r}}}{2^{3 \mathrm{r}}}+\ldots \text { upto } \mathrm{m} \text { terms }\right)$ $=\sum_{\mathrm{r}=0}^{\mathrm{n}}(-1)^{\mathrm{r}} \cdot{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} \frac{1}{2^{\mathrm{r}}}+\sum_{\mathrm{r}=0}^{\mathrm{n}}(-1)^{\mathrm{r}} \cdot{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} \cdot \frac{3^{\mathrm{r}}}{2^{2 \mathrm{r}}}$ $\quad+\sum_{\mathrm{r}=0}^{\mathrm{n}}(-1)^{\mathrm{r}} \cdot{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} \cdot \frac{7^{\mathrm{r}}}{2^{3 \mathrm{r}}}+\ldots$ $=\left(1-\frac{1}{2}\right)^{\mathrm{n}}+\left(1-\frac{3}{4}\right)^{\mathrm{n}}+\left(1-\frac{7}{8}\right)^{\mathrm{n}}+$ $\quad$
...upto $\mathrm{m}$ terms
$=\frac{1}{2^{\mathrm{n}}}+\frac{1}{4^{\mathrm{n}}}+\frac{1}{8^{\mathrm{n}}}+\ldots$ upto $\mathrm{m}$ terms
$=\frac{\frac{1}{2^{\mathrm{n}}}\left(1-\frac{1}{2^{\mathrm{m} \mathrm{n}}}\right)}{\left(1-\frac{1}{2^{\mathrm{n}}}\right)}=\frac{2^{\mathrm{mn}}-1}{2^{\mathrm{mn}}\left(2^{\mathrm{n}}-1\right)}$

Topic: Mathematics, Binomial Theorem, JEE Main
The sum of the slopes of all possible chords of the circle x2+y2=100, which passes through 7,1 and also divide the circumference of this circle into two arcs whose lengths are in the ratio 2:1, is equal to

Options:
A 712
B 127
C -712
D -127

The correct answer is:
712

Solution:
Let, the chord is AB which subtends an angle θ at the centre 0,0
⇒θ+2θ=360o
⇒θ=120o=∠AOB
Let, the distance of O from AB is equal to h
Then, cos⁡60o=h10=12⇒h=5
Let, the equation of chord is y-1x-7=m
⇒mx-y+1-7m=0 whose distance from 0,0 is equal to 5
0-0+1-7m1+m2=5
⇒1-14m+49m2=25+25m2
⇒24m2-14m-24=0⇒m1+m2=1424=712

Topic: Mathematics, Circle, JEE Main
The sum of the solutions of the equation x-2+xx-4+2=0,x>0 is equal to

Options:
A 10
B 9
C 12
D 4

The correct answer is:
10

Solution:
Given,

x-2+xx-4+2=0
⇒x-2+x-4x+2=0
⇒x-2+x2-4x+4=2 (Adding 2 on both sides)
⇒x-22+x-2=2
⇒t2+t-2=0 ; x-2=t ; t>0
⇒t=1or t=-2 ; Rejected because t>0
⇒x-2=1
⇒x=3 or x=1
⇒x=9 or x=1
Hence, the required sum =10.

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2019 (08 Apr Shift 1)
The sum of the square of the modulus of the elements in the set $\{z=\mathrm{a}+\mathrm{ib}: \mathrm{a}, \mathrm{b} \in \mathbf{Z}, z \in \mathbf{C},|z-1| \leq 1,|z-5| \leq|z-5 \mathrm{i}|\}$ is ________

Options:

The correct answer is:
9

Solution:
$\begin{aligned} & |\mathrm{z}-1| \leq 1 \\ & \Rightarrow|(\mathrm{x}-1)+\mathrm{iy}| \leq 1 \mid \\ & \Rightarrow \sqrt{(\mathrm{x}-1)^2+\mathrm{y}^2} \leq 1 \\ & \Rightarrow(\mathrm{x}-1)^2+\mathrm{y}^2 \leq 1 \ldots(1)\end{aligned}$
$\begin{aligned} & \text { Also }|z-5| \leq|z-5 i| \\ & (x-5)^2+y^2 \leq x^2+(y-5)^2 \\ & -10 x \leq-10 y \\ & \Rightarrow x \geq y.....(2)\end{aligned}$
$\begin{aligned} & \text { Solving (1) and (2) } \\ & \begin{array}{l}\Rightarrow(\mathrm{x}-1)^2+\mathrm{x}^2=1 \\ \Rightarrow 2 \mathrm{x}^2-2 \mathrm{x}=0 \\ \Rightarrow \mathrm{x}(\mathrm{x}-1)=0 \\ \Rightarrow \mathrm{x}=0 \text { or } \mathrm{x}=1 \\ \quad \mathrm{y}=0 \text { or } \mathrm{y}=1\end{array}\end{aligned}$
Given $\mathrm{x}, \mathrm{y} \in \mathrm{I}$
Points $(0,0),(1,0),(2,0),(1,1),(1,-1)$ to find
$\begin{aligned}
& \left|z_1\right|^2+\left|z_2\right|^2+\left|z_3\right|^2+\left|z_4\right|^2+\left|z_5\right|^2 \\
& =0+1+4+1+1+1+1=9
\end{aligned}$

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2024 (09 Apr Shift 1)
The sum of the squares of the lengths of the chords intercepted on the circle, x2+y2=16, by the lines, x+y=n, n∈N, where N is the set of all natural numbers is:

Options:
A 210
B 105
C 320
D 160

The correct answer is:
210

Solution:
The length of the perpendicular from centre 0,0 to the line x+y=n is 0+0-n12+I2=n2

Length of intercept =242-n22=64-2n2

⇒ Possible values of n are 1,2,….5

⇒ Sum of the square of lengths of intercepts

=Σn=1564-2n2=64×5-2Σn=15n2

=320-2Σn=15n2=320-110=210

Topic: Mathematics, Circle, JEE Main, JEE Main 2019 (08 Apr Shift 1)
The sum of the total number of bonds between chromium and oxygen atoms in chromate and dichromate ions is ______________.

Options:

The correct answer is:
12

Solution:
Chromate ion
Dichromate ion

The chromate ion (CrO42–) has a tetrahedral shape, whereas dichromate ion has two tetrahedral units with one common oxygen having Cr-O-Cr bond angle of 126°. Both are oxyanions of chromium in the 6+ oxidation state and are moderately strong oxidizing agents.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2020 (09 Jan Shift 2)
The sum of the total number of sigma bonds between chromium and oxygen atoms in chromate and dichromate ions is

Options:

The correct answer is:
12

Solution:
In chromate CrO42- and dichromate Cr2O72- there are 4 and 8 Cr-O bonds are there as follows
Total number of Cr-O bond = 12.

Topic: Chemistry, d and f Block Elements, JEE Main
The sum of the y-intercepts of the tangents drawn from the point -2,-1 to the hyperbola x23-y22=1 is

Options:

The correct answer is:
6

Solution:
Let, the equation of the tangent from -2,-1 is y=mx+3m2-2
â-1=-2m+3m2-2
â2m-12=3m2-2
âm2-4m+3=0
âm=1,3
So, the equations of the tangents are
y+1=x+2 & y+1=3x+2
ây-intercepts of the tangents are 1 & 5
â the sum of y-intercepts =6

Topic: Mathematics, Hyperbola, JEE Main
The sum of total number of lone − pairs of electrons and sp3 hybridized nitrogen atoms in Melamine is

Options:

The correct answer is:
9

Solution:
In Melamine there are 6 number of lone − pairs of electrons and 3 sp3 hybridized nitrogen atoms. All NH2 groups are having sp3 hybridized nitrogen atoms.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
The sum of two forces is 18 N and resultant whose direction is at right angles to the smaller force is 12 N. The magnitude of the two forces are

Options:
A
13, 5

B
12, 6

C
14, 4

D
11, 7

The correct answer is:
13, 5

Solution:
Let two forces be $P$ and $Q$. Given $P+Q=18$ and $P \hat{a}+Q \hat{b}=12 \hat{c} \Rightarrow P \hat{a}-12 \hat{c}=\bar{Q} \hat{b}$
$$
\begin{aligned}
& \Rightarrow P^2+144=Q^2=(18-P)^2 ; \Rightarrow P^2+144=324-36 P+P^2 \\
& \Rightarrow 36 P=180 \Rightarrow P=5 \text { and } Q=13
\end{aligned}
$$
(where $\vec{a}$ and $\vec{b}$ are unit vectors along $P$ and $Q$ ).

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2002
The sum of two forces P→ and Q→ is R→ such that R→=P→. Find the angle between resultant of 2P→ and Q→ and Q→ , ________

Options:

The correct answer is:
90

Solution:
So angle between 2P→+Q→ and Q→ is 90o

Alternate solution

P→+Q→=P→
P2+Q2+2PQcos⁡θ=P2
⇒Q+2PQcos⁡θ=O
⇒cos⁡θ=-Q2P
tan⁡α=2Psin⁡θ2Pcos⁡θ+Q=∞
∵2Pcos⁡θ+Q=0
α=90o

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2020 (07 Jan Shift 2)
The sum of twonumbers is 20 . What are the numbers if the product of the square of one and the cube of the other is maxinmm?

Options:
A 6,14
B 15,5
C 12,8
D 10,10

The correct answer is:
12,8

Solution:
Let the numbers are $\mathrm{x}$ and $\mathrm{y}$. So, $x+y=20$; Let $P=x^{2} y^{3} \quad$ (As given) $=x^{2}(20-x)^{3}$
Differentiating w. r. t. $\mathrm{x}$ $\begin{aligned} \frac{d P}{d x} &=x^{2} \cdot 3(20-x)^{2}(-1)+(20-x)^{3} \cdot 2 x \\ &=(20-x)^{2}\left[-3 x^{2}+40 x-2 x^{2}\right] \\ &=(20-x)^{2}\left[40 x-5 x^{2}\right] \\ \frac{d^{2} P}{d x^{2}} &=(20-x)^{2}[40-10 x]+\left(40 x-5 x^{2}\right) 2(20-x)(-1) \end{aligned}$
$\frac{d p}{d x}=0$ for maxima or minima. So, $(20-x)^{2}\left[40 x-5 x^{2}\right]=0$
$\Rightarrow(20-\mathrm{x})^{2} \times(\mathrm{x})(40-5 \mathrm{x})=0 \Rightarrow \mathrm{x}=20,0,8$
We get, $\left(\frac{\mathrm{d}^{2} \mathrm{P}}{\mathrm{dx}^{2}}\right)_{\mathrm{x}=8} < 0 ;\left(\frac{\mathrm{d}^{2} \mathrm{p}}{\mathrm{dx}^{2}}\right)_{\mathrm{x}=0}>0$ and
$\left(\frac{\mathrm{d}^{2} \mathrm{p}}{\mathrm{dx}^{2}}\right)_{\mathrm{x}=20}=0$
Hence. $\mathrm{P}$ is maximumat $\mathrm{x}=8$
and, Numbers are 12 and 8 .

Topic: Mathematics, Application of Derivatives, JEE Main
The sum of X + Y equal to

Options:

The correct answer is:
5.00

Solution:
Here, X = 2 and Y = 3
So sum of X + Y = 2 + 3 = 5.

Topic: Chemistry, Aldehydes and Ketones, JEE Main
The sum ∑r=110r2+1×r!, is equal to:

Options:
A 11Â Ã11!
B 10 Ã11!
C 11!
D 101Â Ã10!

The correct answer is:
10 Ã11!

Solution:
∑r=110r2+1r!
Let us express the rth term, as a difference involving consecutive factorials, Tr=r2+1+r-r r! =r2+r r!-r-1r!
Therefore,
⇒Tr=rr+1!-r-1r!
Putting r=1,2,3,...,10 and adding, we get ∑r=110rr+1!-r-1r! =10×11!
∴∑r=110r2+1×r!=10×11!

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2016 (10 Apr Online)
The sum âr=150râ2r+250-r is equal to

Options:
A 25250-1
B 50250-1
C 25251-1
D 50251-1

The correct answer is:
50251-1

Solution:
âr=150r2r+250-r
=1⋅2+2⋅22+3⋅23+.â¦â¦..+50⋅250+1⋅249+2⋅248+3⋅247+.â¦..+50⋅20
=200+50+211+49+222+48+.â¦â¦.+24848+2+24949+1+25050+0
=501+2+22+.â¦..+249+250
=50251-12-1=50251-1

Topic: Mathematics, Sequences and Series, JEE Main
The sum to 10 terms of the series 1+21.1+31.12+41.13+.... is

Options:
A 85.12
B 92.5
C 96.75
D 100

The correct answer is:
100

Solution:
Let x=1.1
S=1+2x+3x2+........10x9 and

xS=x+2x2+........9x9+10x10
Subtracting, we get,

S1-x=1+x+x2+.......+x9-10x10
=x10-1x-1-10x10

⇒-S=x10-1x-12-10x10x-1
⇒-S=1.110-10.12-101.1100.1=100×1.110-100-100×1.110=-100
⇒S=100.

Topic: Mathematics, Sequences and Series, JEE Main
The sum to 20 terms of the series 2·22-32+2·42-52+2·62-............ is equal to__________.

Options:

The correct answer is:
1310

Solution:
Given,

2·22-32+2·42-52+2·62-............20 terms,

=∑r=1102·(2r)2-(2r+1)2

=∑r=1108r2-4r2-4r-1

=∑r=1104r2-4r-1

=4∑r=110r2-4∑r=110r-∑r=1101

=4·10·11·216-410·112-10

=44·35-220-10=1540-230=1310

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2023 (13 Apr Shift 1)
The sum to infinite terms of the arithmetic-geometric progression 3, 4, 4, 329, .... is equal to

Options:
A 27
B 30
C 24
D 25

The correct answer is:
27

Solution:
Let the A.G.P. be 3, 3+dr, 3+2dr2....
So, 3+dr=4 and 3+2dr2=4
3+d2r23+2dr2=164
⇒9+d2+6d=12+8d
⇒d2-2d-3=0
⇒d+1d-3=0⇒d=-1,3
r=43+d⇒r=43-1,43+3=2,23
But r<1⇒d=3 and r=23
S∞=a1-r+dr1-r2=31-23+3×231-232
=9+18=27

Topic: Mathematics, Sequences and Series, JEE Main
The sum to infinity of the following series $\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\ldots \ldots \ldots .$. shall be

Options:
A $\infty$
B $1$
C $0$
D None of these

The correct answer is:
$1$

Solution:
Given series
$\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\ldots=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\ldots$
Obviously the sum of series upto infinity is 1.

Topic: Mathematics, Sequences and Series, JEE Main
The sum to infinity of the series 1+45+752+1053+… is

Options:
A 1625
B 115
C 3516
D 811

The correct answer is:
3516

Solution:
S=1+45+752+1053+…+∞ ... (1)
S5=0+15+452+753+…+∞ ... (2)
On subtracting both the eq (1) and (2), we get,
45S=1+35+352+…+∞
S=3516

Topic: Mathematics, Sequences and Series, JEE Main
The sum to $(n+1)$ terms of the following series $\frac{C_0}{2}-\frac{C_1}{3}+\frac{C_2}{4}-\frac{C_3}{5}+$ is

Options:
A $\frac{1}{n+1}$
B $\frac{1}{n+2}$
C $\frac{1}{n(n+1)}$
D None of these

The correct answer is:
None of these

Solution:
$\begin{aligned} & (1-x)^n=C_0-C_1 x+C_2 x^2-C_3 x^3+\ldots . \\ & \Rightarrow x(1-x)^n=C_0 x-C_1 x^2+C_2 x^3-C_3 x^4+\ldots . .\end{aligned}$
$\Rightarrow \int_0^1 x(1-x)^n d x=\int_0^1\left(C_0 x-C_1 x^2+C_2 x^3 \ldots\right) d x...(i)$
The integral on the LHS
$=\int_1^0(1-t) t^n(-d t)$
by putting $1-x=t$
$=\int_0^1\left(t^n-t^{n+1}\right) d t=\frac{1}{n+1}-\frac{1}{n+2}$
Whereas the integral on the RHS of (i)
$\begin{aligned} & =\left[\frac{C_0 x^2}{2}-\frac{C_1 x^3}{3}+\frac{C_2 x^4}{4}-\ldots .\right]=\frac{C_0}{2}-\frac{C_1}{3}+\frac{C_2}{4}-\ldots \\
& \therefore \frac{C_0}{2}-\frac{C_1}{3}+\frac{C_2}{4}-\ldots \text { to }(n+1) \text { terms } \\
& =\frac{1}{n+1}-\frac{1}{n+2}=\frac{1}{(n+1)(n+2)} \end{aligned}$

Topic: Mathematics, Binomial Theorem, JEE Main
The sum to n terms of 11.3+21.3.5+31.3.5.7+41.3.5.7.9+........ is

Options:
A 121+11.3.5........2n+1
B 121-12.4.6........2n
C 121-11.3.5........2n+1
D None of these

The correct answer is:
121-11.3.5........2n+1

Solution:
tn=n1.3.5. .........2n+1
= 12. 2n+1-11.3.5. .........2n+1
=1211.3.5......2n-1-11.3.5.......2n+1
=12Tn-1- Tn
∴2tn=Tn-1- Tn  …1 where Tn=11.3.5.......2n+1
∴  2 S n  =  ∑ n=2 n 2 t n +2 t 1
⇒ 2 S n  − 2 t 1  = ( T 1  −  T 2 )+( T 2  −  T 3 )+........+( T n−1 − T n )
⇒ 2 S n  − 2 t 1  =  T 1 − T n
⇒ 2 S n  = 2 t 1  +  T 1 − T n                = 2× 1 3 + 1 1×3 − 1 1×3×5.....( 2n+1 )
⇒ Sn = 121−11×3×5.......(2n+1)

Topic: Mathematics, Sequences and Series, JEE Main
The sum to $n$ terms of $(2 n-1)+2(2 n-3)+3(2 n-5)+\ldots . \cdot$ is

Options:
A $(n+1)(n+2)(n+3) / 6$
B $n(n+1)(n+2) / 6$
C $n(n+1)(2 n+3)$
D $n(n+1)(2 n+1) / 6$

The correct answer is:
$n(n+1)(2 n+1) / 6$

Solution:
$S=(2 n-1)+2(2 n-3)+3(2 n-5)+\ldots$
$S=[2 n+2.2 n+3.2 n+\ldots \ldots .+n .2 n]-$ $[1+2 \cdot 3+3 \cdot5+\ldots .+n \cdot(2 n-1)]$
Let, $S_1=2 n(1+2+3+\ldots .+n)=\frac{2 n \cdot n(n+1)}{2}=n^2(n+1)$
and $S_2=1+2 \cdot 3+3 \cdot 5+\ldots . .+n \cdot(2 n-1)$
$T_n=n(2 n-1)=2 n^2-n$
$\therefore S_2=\Sigma\left(2 n^2-n\right)=2 \Sigma\left(n^2\right)-\Sigma(n)$
$=\frac{2 n(n+1)(2 n+1)}{6}-\frac{n(n+1)}{2}$
$\therefore S=S_1-S_2=n^2(n+1)-\frac{2 n(n+1)(2 n+1)}{6}+\frac{n(n+1)}{2}$
$\therefore S=S_1-S_2=n^2(n+1)-\frac{2 n(n+1)(2 n+1)}{6}+\frac{n(n+1)}{2}$
$=n(n+1)\left[n-\frac{2 n+1}{3}+\frac{1}{2}\right]$
$=n(n+1)\left[\frac{6 n-4 n-2+3}{6}\right]=\frac{n(n+1)(2 n+1)}{6}$

Topic: Mathematics, Sequences and Series, JEE Main
The sum to $\mathrm{n}$ terms of the series

$\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+\ldots \ldots \ldots \ldots$ is

Options:
A $n-1-2^{-n}$
B 1
C $n-1+2^{-n}$
D $1+2^{-n}$

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

Solution:
$\begin{aligned} \frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+\ldots \\=&\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{8}\right)+\left(1-\frac{1}{16}\right)+\ldots . \\=& \mathrm{n}-\frac{\frac{1}{2}\left\{1-\frac{1}{2^{\mathrm{n}}}\right\}}{1-\frac{1}{2}}=\mathrm{n}-1+2^{-\mathrm{n}} \end{aligned}$

Topic: Mathematics, Sequences and Series, JEE Main
The sum to $n$ terms of the series $2^2+4^2+6^2+$\ldots\ldotsis

Options:
A $\frac{n(n+1)(2 n+1)}{3}$
B $\frac{2 n(n+1)(2 n+1)}{3}$
C $\frac{n(n+1)(2 n+1)}{6}$
D $\frac{n(n+1)(2 n+1)}{9}$

The correct answer is:
$\frac{2 n(n+1)(2 n+1)}{3}$

Solution:
$2^2+4^2+6^2+\ldots \ldots \ldots+(2 n)^2$ $=2^2\left(1^2+2^2+3^2+\ldots \ldots . .+n^2\right)$
$=\frac{4 n(n+1)(2 n+1)}{6}=\frac{2 n(n+1)(2 n+1)}{3}$

Topic: Mathematics, Sequences and Series, JEE Main
The sum to \( n \) terms of the series \( 1+3+7+15+ \) is

Options:
A \( 2^{n+1}-n \)
B \( 2^{n}-n-2 \)
C \( 2^{\mathrm{n}}-\mathrm{n} \)
D \( 2^{n+1}-n-2 \)

The correct answer is:
\( 2^{n+1}-n-2 \)

Solution:
S = 1 + 3 + 7 + 15 + 31 + ........
S = 1 + 3 + 7 + 15 + ..........
---------------------------------------------------
0 = (1 + 2 + 4 + 8 + 16 + ......... n terms )- Tn (on subtracting)
â´ Tâ¡ n = 1 + 2 + 4 + 8 + ....â¡ to n terms = 2 n - 1 2 - 1
â´ Sâ¡ n = Î£ n = 1 n Tâ¡ n = Î£ n = n 2 n - 1 = Î£ n = 1 n 2 n - Î£ n = 1 n 1
= 2 Ã 2 n - 1 2 - 1 - n = 2 n + 1 - n - 2

Topic: Mathematics, Sequences and Series, JEE Main
The sum to \( n \) terms of the series \( 1.3^{2}+2.5^{2}+3.7^{2}+\ldots \) is

Options:
A \( \frac{n}{6}(n+1)\left(6 n^{2}+14 n+7\right) \)
B \( \frac{n}{6}(n+1)(2 n+1)(3 n+1) \)
C \( 4 n^{3}+4 n^{2}+n \)
D None of the above

The correct answer is:
\( \frac{n}{6}(n+1)\left(6 n^{2}+14 n+7\right) \)

Solution:
Given series is 1.32+2.52+3.72+ ....∞.

This is an arithmetic-geometric series whose nth term is equal to

Tn=n2n+12=4n3+4n2+n

∴ Sn=∑1nTn=∑1n4n3+4n2+n

=4∑1nn3+4∑1nn2+∑1nn

=4n2n+12+46nn+12n+1+n2n+1

=nn+1n2+n+462n+1+12

=n6n+16n2+14n+7

Topic: Mathematics, Sequences and Series, JEE Main
The sum to the infinite terms of the series 532⋅72+972⋅112+13112⋅152+.... is

Options:
A 18
B 136
C 154
D 172

The correct answer is:
172

Solution:
Tk=5+k-143+k-1427+k-142
=4k+14k-124k+32
=1814k-12-14k+32
Sn=T1+T2+T3+....+Tn
=18132-172+172-1112+1112-1152+....+14n-12-14n+32
=18132-14n+32
S∞=limn→∞Sn=1819-0=172

Topic: Mathematics, Sequences and Series, JEE Main
The sum to the infinity of the series $1+\frac{2}{3}+\frac{6}{3^2}+\frac{10}{3^3}+\frac{14}{3^4}+\ldots \ldots$ is

Options:
A
2

B
3

C
4

D
6

The correct answer is:
3

Solution:
Let $S=1+\frac{2}{3}+\frac{6}{3^2}+\frac{10}{3^3}+\frac{14}{3^4}+\ldots$.
$\frac{1}{3} S=\frac{1}{3}+\frac{2}{3^2}+\frac{6}{3^3}+\frac{10}{3^4}+\ldots$
Dividing (1) \& (2)
$$
\begin{aligned}
& S\left(1-\frac{1}{3}\right)=1+\frac{1}{3}+\frac{4}{3^2}+\frac{4}{3^3}+\frac{4}{3^4}+\ldots . \\
& \frac{2}{3} S=\frac{4}{3}+\frac{4}{3^2}\left(1+\frac{1}{3}+\frac{1}{3^2}+\ldots . .\right) \Rightarrow \frac{2}{3} S=\frac{4}{3}+\frac{4}{3^2}\left(\frac{1}{1-\frac{1}{3}}\right)=\frac{4}{3}+\frac{4}{3^2} \frac{3}{2}=\frac{4}{3}+\frac{2}{3}=\frac{6}{2} \Rightarrow \frac{2}{3} S=\frac{6}{3} \Rightarrow S=3
\end{aligned}
$$

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2009
The sum up to 60 terms of 312+512+22+712+22+32+.... is equal to

Options:
A 24061
B 18017
C 36061
D 10017

The correct answer is:
36061

Solution:
TK=2K+112+22+....+K2=2K+1K K+1 (2K+1)6

=6KK+1=61K-1K+1
∑K=160TK=61-12+12-13+......+160-161
=61-161=6×6061=36061

Topic: Mathematics, Sequences and Series, JEE Main
The sum (upto two decimal places) of the infinite series 717+77172+777173+....... is

Options:
A 1.06
B 2.06
C 3.06
D 4.06

The correct answer is:
1.06

Solution:
S=717+77172+777173+.......

S17=7172+77173+777174+........

Subtracting both the above results, we get,

1617S=717+70172+700173+.......
Infinite G.P.with r =1017
1617S=7171-1017=717717=1
S=1716

Topic: Mathematics, Sequences and Series, JEE Main
The summation of three sinusoidal waves is equal to

Options:
A AM Waves
B FM Waves
C Both (1) and (2)
D Neither of these

The correct answer is:
AM Waves

Solution:
The summation of three sinusoidal waves is equal to AM Waves.

Topic: Physics, Communication System, JEE Main
The sums of $n$ terms of two a arithmatic series are in the ratio $2 n+3: 6 n+5$, then the ratio of their $13^{\text {th }}$ terms is

Options:
A $53: 155$
B $27: 77$
C $29: 83$
D $31: 89$

The correct answer is:
$53: 155$

Solution:
We have
$\frac{S_{n_1}}{S_{n_2}}=\frac{2 n+3}{6 n+5}$
$\Rightarrow \frac{\frac{n}{2}\left[2 a_1+(n-1) d_1\right]}{\frac{n}{2}\left[2 a_2+(n-1) d_2\right]}$ $=\frac{2 n+3}{6 n+5}$
$\Rightarrow \frac{2\left[a_1+\left(\frac{n-1}{2}\right) d_1\right]}{2\left[a_2+\left(\frac{n-1}{2}\right) d_2\right]}$ $=\frac{2 n+3}{6 n+5}$
$\Rightarrow \frac{a_1+\left(\frac{n-1}{2}\right) d_1}{a_2+\left(\frac{n-1}{2}\right) d_2}$ $=\frac{2 n+3}{6 n+5}$
Put $n=25$ then $\frac{a_1+12 d_1}{a_2+12 d_2}$ $\frac{a_1+12 d_1}{a_2+12 d_2}=\frac{2(25)+3}{6(25)+3}$ $\Rightarrow \frac{T_{13_1}}{T_{13_2}}=\frac{53}{155}$

Topic: Mathematics, Sequences and Series, JEE Main
The sun is a hot plasma (ionised matter) with its inner core at a temperature exceeding $10^7 \mathrm{~K}$, and its outer surface at a temperature of about $6000 \mathrm{~K}$. At such high temperatures, no substance remains in a solid or liquid phase. In what range do you expect the mass density of the Sun to be, in the range of densities of solids, liquids or gases? Check if your guess is correct from the following data: mass of $S u n=2 \times 10^{30} \mathrm{~kg}$, radius of the Sun $=7 \times 10^8 \mathrm{~m}$.

Options:

None

Solution:
$\mathrm{M}=2 \times 1030 \mathrm{~kg}, \quad \mathrm{R}=7 \times 108 \mathrm{~m}$
Density, $\rho=\frac{\text { mass }}{\text { volume }}=\frac{\mathrm{M}}{\frac{4}{3} \pi \mathrm{R}^3}=\frac{3 \mathrm{M}}{4 \pi \mathrm{R}^3}$ $=\frac{3 \times 2 \times 10^{30}}{4 \times 3.14 \times\left(7 \times 10^8\right)^3}=1.392 \times 103 \mathrm{~kg} / \mathrm{m}^3$. which is the order of density of solids and liquids not gases.
High density of Sun is because of inward gravitational attractions or the outer layers of the Sun.

Topic: Physics, Units and Dimensions
The supply voltage to a room is 120 V. The resistance of the lead wires is 6 Ω . A 60 W bulb is already switched on. What is the decrease of voltage across the bulb, when a 240 W heater is switched on in parallel to the bulb?

Options:
A 13.3 V
B 10.4 V
C zero V
D 2.9 V

The correct answer is:
10.4 V

Solution:
Resistance of bulb, Rb=V2Pb=120×12060=240 Ω
Resistance of heater, Rh=V2Ph=120×120240=60 Ω

Voltage across bulb before the heater is not connected, V1=V×RbRb+6=120×240 V246=117.07 V

Now the bulb and heater are connected in parallel.

Resistance of the combination is Rnet=240×60240+60=48 Ω
Voltage across bulb after the heater is switched on,
V2=V×RnetRnet+6=120×4848+6=106.66 V

The decrease in the voltage is V1-V2=117.07-106.66≃10.4 V

Note: Here supply voltage is taken as rated voltage.

Topic: Physics, Current Electricity, JEE Main, JEE Main 2013 (07 Apr)