Question

Let the general term of a series be $(2k-1)(2k)$ $(2k+1),k=1,2,\mathrm{3...},n.$ If the sum of first. $n$ terms is $24090,$ then $n$ is equal to

• A. $8$
• B. $9$
• C. $10$
• D. $11$

Answer 1

Answer: (C)

Given, general term of a series $T_k=(2k-1)(2k)(2k+1).$
$\Rightarrow$ $T_k=\{{(2k)}^2-{(1)}^2\}(2k)$
$\Rightarrow$ $T_k=(4k^2-1)(2k)$
$\Rightarrow$ $T_k=8k^3-2k$
$\Rightarrow$ $T_n=8n^3-2n$ Then, $S_n=\Sigma T_n$
$=\Sigma (8n^3-2\Sigma n)$
$\Rightarrow$ $S_n=8\Sigma n^3-2\Sigma n$
$=8{\left(\frac{n(n+1)}{2}\right)}^2-2\frac{n(n+1)}{2}$
$=8\frac{n^2{(n+1)}^2}{4}-n(n+1)$ $2n^2{(n+1)}^2-n(n+1)$ But given, $S_n=24090$
$\therefore$ $2n^2{(n+1)}^2-n(n+1)=24090$
$\Rightarrow$ $2n^2(n^2+2n+1)-n(n+1)=24090$
$\Rightarrow$ $2(n^4+2n^3+n^2)-n^2-n=24090$
$\Rightarrow$ $2n^4+4n^3+2n^2-n^2-n=24090$
$\Rightarrow$ $2n^4+4n^3+n^2-n=24090$
$\Rightarrow$ $n(2n^3+4n^2+n-1)=24090$
Now, $n=10$ satisfy this equation. So, $n=10$