Exp. (b)Statement IS=(1)+(1+2+4)+(4+6+9)+(9+12+16)+…+(361+380+400)S=(0+0+1)+(1+2+4)+(4+6+9)+(9+12+16)+…+(361+380+400)Now, we can clearly observe the first elements in each bracket.In second bracket, the first element is 1=12In third bracket, the first element is 4=22In fourth bracket, the first element is 9=32In last bracket, the first element is 361=192Hence, we can conclude that there are 20 brackets in all.Also, in each of the brackets, there are 3 terms out of which the first and last terms are perfect squares of consecutive integers and the middle term is their product.⇒ The general term of the series isTr​=(r−1)2+(r−1)r+(r)2⇒ The sum of the n terms of the series isSn​=∑r=1n​[(r−1)2+(r−1)r+(r)2]Sn​=∑r=1n​[r−(r−1)r3−(r−1)3​][∵a3−b3=(a−b)(a2+ab+b2)]⇒ Sn​=∑r=1n​[r3−(r−1)3]Now, let Sn​=∑k=1n​[k3−(k−1)3]On substituting the value of k, we getSn​=∑k=1n​[k3−(k−1)3]=(13−03)+(23−13)+(33−23)+…+[n3−(n−1)3]On rearranging the terms, we get Sn​=−03+(13−13)+(23−23)+(33−33)+…+[(n−1)3−(n−1)3]+n3⇒Sn​=n3Since, the number of terms is 20 , hence substituting n=20, we getS20​=8000Hence, Statement I is correct.Statement II We have already proved in the Statement I, thatSn​=∑k=1n​[k3−(k−1)3]=n3Hence, Statement II is also correct and is a correct explanation of Statement I.Caution If you have solved the series in Statement 1 , in the following way, then you may get option (b) as the correct answer. The sum of the nterms of the series is Sn​=∑r=1n​[(r−1)2+(r−1)r+(r)2]⇒Sn​=∑r=0n​(3r2−3r+1)⇒Sn​=3∑r=1n​r2−3∑r=1n​r+∑r=1n​1⇒Sn​=36n(n+1)(2n+1)​−32r(n+1)​+n