Question:
If \(1+(1-2^2\cdot 1)+(1-4^2\cdot 3)+(1-6^2\cdot 5)+\ldots+(1-20^2\cdot 19)=\alpha-220\beta\), then an ordered pair \((\alpha,\beta)\) is equal to :
If \(1+(1-2^2\cdot 1)+(1-4^2\cdot 3)+(1-6^2\cdot 5)+\ldots+(1-20^2\cdot 19)=\alpha-220\beta\), then an ordered pair \((\alpha,\beta)\) is equal to :
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The general term after the first is \(1-(2k)^2(2k-1)=1-8k^3+4k^2\) for \(k=1,\dots,10\); use the sums of \(k^2\) and \(k^3\).
Updated On: Nov 5, 2026
- (10, 97)
- (10, 103)
- (11, 103)
- (11, 97)
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The Correct Option is C
Solution and Explanation
The sum is \(1+\sum_{k=1}^{10}\left[1-(2k)^2(2k-1)\right]=1+10-\sum_{k=1}^{10}(8k^3-4k^2)\).
\(\sum_{k=1}^{10}k^3=\left(\frac{10\cdot 11}{2}\right)^2=3025\) and \(\sum_{k=1}^{10}k^2=\frac{10\cdot 11\cdot 21}{6}=385\).
So the sum \(=11-(8\cdot 3025-4\cdot 385)=11-(24200-1540)=11-22660\).
Now \(22660=220\cdot 103\), so the sum \(=11-220\cdot 103\), giving \(\alpha=11,\ \beta=103\).
Answer: (3) (11, 103)
\(\sum_{k=1}^{10}k^3=\left(\frac{10\cdot 11}{2}\right)^2=3025\) and \(\sum_{k=1}^{10}k^2=\frac{10\cdot 11\cdot 21}{6}=385\).
So the sum \(=11-(8\cdot 3025-4\cdot 385)=11-(24200-1540)=11-22660\).
Now \(22660=220\cdot 103\), so the sum \(=11-220\cdot 103\), giving \(\alpha=11,\ \beta=103\).
Answer: (3) (11, 103)
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