JEE Advanced 2013 · Paper 1 · Mathematics Q51

MathematicsMultiple correct+4 / -1Question 51 of 60

Let $S_n=\displaystyle\sum_{k=1}^{4n}(-1)^{\frac{k(k+1)}2}k^2$ . Then $S_n$ can take value(s)

1056

1088

1120

1332

Answer and worked solution

Official answer: A, D

Sign pattern. $\frac{k(k+1)}2$ for $k=1,2,3,4$ is 1, 3, 6, 10: signs −, −, +, +, repeating every 4 terms.

Each block of four ( $k=4m-3,\ldots,4m$ ):
$(4m)^2-(4m-2)^2+(4m-1)^2-(4m-3)^2=2(8m-2)+2(8m-4)=32m-12.$

Sum. $S_n=16n(n+1)-12n=16n^2+4n$ . $n=8$ : 1056; $n=9$ : 1332. 1088 and 1120 are not of this form.

(A), (D)

Topics:Special series sums

JEE Advanced 2013 · Paper 1, Mathematics question 51