JEE Advanced 2013 Paper 1

Sn =

4n

∑

k=1

(−1)

k(k+1)

2

k2
= −12−22+32+42−52−62+72+82+.....

= 2[4+6+12+14+20+22+.....]→ 2n terms
= 2[(4 + 12 + 20 ....) + (6 + 14 + 22 .....)] → n terms
= 2[

n

2

(4×2 +(n-1)8)

n

2

(2×6+(n+1)8)]
= 2[n(4+4n - 4) + n(6 + 4n - 4)]
= 2(4n2 + (4n + 2)n)
= 2(8n2 + 2n)
= 4n(4n + 1)
(A) 1056 = 32 x 33 for n=8 ,
(B)1088 = 32 x 34,
(C)1120 = 32 x 35,
(D)1332 = 36 x 37 for n = 9