Prove the following
Question:

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

  1. (1) $(10,97)$

  2. (2) $(11,103)$

  3. (3) $(10,103)$

  4. (4) $(11,97)$


Correct Option: , 2

Solution:

The given series is

$1+\left(1-2^{2} \cdot 1\right)+\left(1-4^{2} \cdot 3\right)+\left(1-6^{2} \cdot 5\right)+\ldots\left(1-20^{2} \cdot 19\right)$

$S=1+\sum_{r=1}^{10}\left[1-(2 r)^{2}(2 r-1)\right]$

$=1+\sum_{r=1}^{10}\left(1-8 r^{3}+4 r^{2}\right)=1+10-\sum_{r=1}^{10}\left(8 r^{3}-4 r^{2}\right)$

$=11-8\left(\frac{10 \times 11}{2}\right)^{2}+4 \times\left(\frac{10 \times 11 \times 21}{6}\right)$

$=11-2 \times(110)^{2}+4 \times 55 \times 7$

$=11-220(110-7)$

$=11-220 \times 103=\alpha-220 \beta$

$\Rightarrow \alpha=11, \beta=103$

$\therefore(\alpha, \beta)=(11,103)$

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