Question

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

• A. $16n^2+4n.$
• B. $16n^2+3n.$
• C. $18n^2+4n.$
• D. $16n^2+5n.$

Answer 1

The correct option is A $16n^2+4n.$

$S=-1^2-2^2+3^2+4^2....................+{(4n-1)}^2+\left({4n}^2\right)\to \left(1\right)$

$\Rightarrow S=4n^2+{(4n-1)}^2-{(4n-2)}^2-{(4n-3)}^2+...........-2^2-1^2\to \left(II\right)$ ( in reverse order )

Adding them -

$\Rightarrow 2S=(4n-1)(4n+1)+(4n-3)(4n+1)-(4n-5)(4n+1).......-(3-4n)(4n+1)-1(1-4n)(4n+1)$

Factoring out $(4n+1)$ all that's left is to complete the following sum ( let's call it A )

$\Rightarrow (4n-1)+(4n-3)-(4n-5)-(4n-7)+.........(3-4n)-(1-4n)=A$

But, sum of each $4$ consecutive numbers of A is equal to $8,$ because if we call the first number in each group of for $x,$ then we have

$\Rightarrow x+(x-2)-(x-4)-(x-6)=8$

As there are $4n$ numbers in the sum that is equal to A, there are n groups of $4$ and $\therefore A=8n$

Therefore, we have

$\Rightarrow (4n+1)A=2S$

$\Rightarrow 8n(4n+1)=2S$

and finally,

$\Rightarrow S=4n(4n+1)=16n^2+4n$

Hence, the answer is $16n^2+4n.$