Question

If ${S_n} = \sum\limits_{k = 1}^{4n} {{{( - 1)}^{\dfrac{{k(k + 1)}}{2}}}} {k^2}$ then, ${S_n}$ can take values
A. $1056$
B. $1088$
C. $1120$
D. $1332$

Answer 1

Hint: The given expression means the sum of all the ${( - 1)^{\dfrac{{k(k + 1)}}{2}}}{k^2}$ terms where $k$ takes the values from $1$ to $4n$ . We can also write $\sum\limits_{k = a}^b {{{( - 1)}^{\dfrac{{k(k + 1)}}{2}}}} {k^2}$ where $k$ takes the values from $a$ to $b$ , $a$ being the lower limit and $b$ be the upper limit.

Complete step by step solution: 
We have ${S_n} = \sum\limits_{k = 1}^{4n} {{{( - 1)}^{\dfrac{{k(k + 1)}}{2}}}} {k^2}$ .
If we expand the summation from $k = 1$ to $k = 4n$ we will get the following sequence,
$ = - {1^2} - {2^2} + {3^2} + {4^2} - {5^2} - {6^2} + {7^2} + \ldots + {(4n)^2}$
Now separating the even and odd terms in the sequence, we get
$ = ( - {1^2} + {3^2} - {5^2} + \ldots + {(4n - 1)^2}) + ( - {2^2} + {4^2} - {6^2} + \ldots + {(4n)^2})$
On further solving we have
$ = 2(4 + 12 + \ldots + (8n - 4)) + 2(6 + 14 + \ldots (8n - 2))$
This consists of two AP sequences of n-terms
$2(4 + 12 + \ldots + (8n - 4))$ and $2(6 + 14 + \ldots + (8n - 2))$
$ = n(8n) + n(8n + 4)$
Since it is in AP, we can apply the formula for the sum of n-terms ${S_n} = \dfrac{n}{2}[2a + (n - 1)d]$ in both the sequences.
After evaluating the sum of n-terms, we get
$ = 4n(4n + 1)$
We have sum $1056$ for $n = 8$ and sum $1332$ for $n = 9$
Hence, we have two correct options: (A) and (D).

Note: To write a sum in sigma notation, try to find a formula involving a variable $k$ where the first term can be obtained by setting $k = 1$ , the second term by $k = 2$ , and so on. Regarding AP sum we have to see if the sum of n-terms ${S_n}$ is given, then the general term ${T_n} = {S_n} - {S_{n - 1}}$ , where ${S_{n - 1}}$ is sum of $(n - 1)$ terms of the arithmetic progression.