If 1+(1–22.1)+(1–42.3)+(1-62.5)+......+(1-202.19)= $\alpha$ - 220$\beta$,
then an ordered pair $\left( {\alpha ,\beta } \right)$ is equal to:
Solution
$1 + (1 - {2^2}.1) + (1 - {4^2}.3) + (1 - {6^2}.5) + ....(1 - {20^2}.19)$<br><br>$$S = 1 + \sum\limits_{r = 1}^{10} {\left[ {1 - {{(2r)}^2}(2r - 1)} \right] = 1 + \sum\limits_{r = 1}^{10} {\left( {1 - 8{r^3} + 4{r^2}} \right)} = 1 + 10 - } \sum\limits_{r = 1}^{10} {\left( {8{r^3} - 4{r^2}} \right)} $$<br><br>$$= 11 - 8{\left( {{{10 \times 11} \over 2}} \right)^2} + 4 \times \left( {{{10 \times 11 \times 21} \over 6}} \right)$$<br><br>$= 11 - 2 \times {(110)^2} + 4 \times 55 \times 7$<br><br>$= 11 - 220(110 - 7)$<br><br>$= 11 - 220 \times 103 = \alpha - 220\beta$
<br><br>$\Rightarrow \alpha = 11,\beta = 103$
About this question
Subject: Mathematics · Chapter: Sequences and Series · Topic: Arithmetic Progression
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