Question

If $1+(1-2^2\times 1)+(1-4^2\times 3)+(1-6^2\times 5)+......+(1-{20}^2\times 19)=\alpha -220\beta$, then an ordered pair $(\alpha ,\beta )$ is equal to

• A. $(10,97)$
• B. $(11,103)$
• C. $(11,97)$
• D. $(10,103)$

Answer 1

The correct option is B

$(11,103)$

Explanation for the correct option:

To find the value of $(\alpha ,\beta )$

Step 1: Find the general term of given series and then find the sum of series.

The given equation is $1+(1-2^2\times 1)+(1-4^2\times 3)+(1-6^2\times 5)+......+(1-{20}^2\times 19)=\alpha -220\beta$

Let $T_n$be the general term and $S_n$ be the sum of n terms of the given series where $n$is the positive integer.

$\begin{array}{rl}{T}_n& =1-(2n{)}^2(2n-1)\\ & =1-4n^2(2n-1)\\ & =1-8n^3+4n^2\\ S_n& =n-\sum _{n=1}^{10}8n^3+\sum _{n=1}^{10}4n^2\\ & =n-8\times {\left(\frac{n(n+1)}{2}\right)}^2+4\frac{n(n+1)(2n+1)}{6}\end{array}$

Step 2: Put $n=10$ in $S_n$

$\begin{array}{rl}{S}_{10}& =10-2\times 100\times 121+\frac{2}{3}\times 10\times 11\times 21\\ & =10-24200+1540\\ & =10-22660\\ & =11-220\times 103\\ \alpha –220\beta & =11-220\times 103\end{array}$

$\alpha =11,\beta =103$

Hence $(\alpha ,\beta )=(11,103)$ . Hence option$\left(B\right)$ is the correct option.