Question

it has multichoices jee adv 2015

Answer 1

Answer:

${ \sf{ \pink{hey \: there \: here \: is \: your \: answer}}} \\ { \sf{ s_n = \sum_{k = 1} ^{4n} ( - 1)^{ \frac{k(k + 1)}{2} } {k}^{2} \implies \: a)1056 \: d)1332 }}$

Answer 2

Question:

$S_{n}=\sum\limits_{k=1}^{4n} {( - 1)}^{\frac{k(k + 1)}{2} } {k}^{2}$, then $S_{n}$ can take the values

(a) 1056

(b) 1088

(c) 1120

(d) 1332

Answer:

$\large\boxed{\sf{(a)1056\;\;and\;\;(d)1332}}$

Step-by-step explanation:

Given that,

$S_{n}=\sum\limits_{k=1}^{4n} {( - 1)}^{\frac{k(k + 1)}{2} } {k}^{2}$

Putting the limits and solving further , we get,

$= - {1}^{2} - {2}^{2} + {3}^{2} + {4}^{2} - {5}^{2} - {6}^{2} + {7}^{2} + {8}^{2} - ....... \\ \\ = ( {3}^{2} - {1}^{2} ) + ( {4}^{2} - {2}^{2} ) + ( {7}^{2} - {5}^{2} ) + ( {8}^{2} - {6}^{2} ) + ..... \\ \\ = 8 + 12 + 24 + ........ \\ \\ =2 \left[(4 + 12 + 20 + ....) + (6 + 14 + 22 + .....)\right] \\ \\ = 2\left[n(4 + 4n - 4) + n(6 + 4n - 4)\right] \\ \\ = 2\left[4 {n}^{2} +n (4n + 2) \right] \\ \\ = 2(8 {n}^{2} + 2n) \\ \\ = 4n(4n + 1)$

Now, for the given options, we have,

(a) 1056 = 32 × 33

(b) 1088 = 32 × 34

(c) 1120 = 32 × 35

(d) 1332 = 36 × 37

Here, only options (a) and (d) satisfy for the following conditon.

For (a), n = 8

For (d), n = 9

Hence, the correct options are (a)1056 and (d)1332