JEE Mains · Maths · STD 11 - 8. sequence and series
Let \(\alpha=1^2+4^2+8^2+13^2+19^2+26^2+\ldots\) upto 10 terms and \(\beta=\sum_{n=1}^{10} n^4\). If \(4 \alpha-\beta=55 k+40\), then \(\mathrm{k}\) is equal to ...........
- A \(456\)
- B \(353\)
- C \(468\)
- D \(435\)
Answer & Solution
Correct Answer
(B) \(353\)
Step-by-step Solution
Detailed explanation
\(\alpha=1^2+4^2+8^2 \ldots . \) \(t_n=a^2+b n+c\) \( 1=a+b+c\) \( 4=4 a+2 b+c \) \( 8=9 a+3 b+c\) On solving we get, \(\mathrm{a}=\frac{1}{2}, \mathrm{~b}=\frac{3}{2}, \mathrm{c}=-1\) \( \alpha=\sum_{n=1}^{10}\left(\frac{n^2}{2}+\frac{3 n}{2}-1\right)^2 \)…
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