Let = 1 2 + 4 2 + 8 2 + 13 2 + 19 2 + 26 2 + … upto 10 terms and = ? n = 1 10 n 4 . If 55 k + 40 , then k is equal to ______ . [2024]

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Solution

Q.
Let $α = 1^{2} + 4^{2} + 8^{2} + 13^{2} + 19^{2} + 26^{2} + \dots$ upto 10 terms and $β = \sum_{n = 1}^{10} n^{4} .$ If $4 α - β = 55 k + 40,$ then $k$ is equal to ______ . [2024]

Ans.
(353)

$α = 1^{2} + 4^{2} + 8^{2} + 13^{2} \dots$ upto 10 terms, and $β = 1^{4} + 2^{4} + 3^{4} + \dots$ upto 10 terms.

Let us find general terms for $a$

$S_{n} = 1 + 4 + 8 + . . . + a_{n}$

Again $S_{n} =$ $1 + 4 + . . . + a_{n - 1} + a_{n}$

________________________________

$O = 1 + 3 + 4 + . . . (n - 1) t e r m - a_{n}$

$\Rightarrow a_{n} = 1 + 3 + 4 + \dots$

$\Rightarrow a_{n} = 1 + (3 + 4 + 5 + \dots + (n - 1) terms)$

$\Rightarrow a_{n} = 1 + \frac{n - 1}{2} (4 + n) = \frac{2 + (n - 1) (4 + n)}{2}$

$\frac{n^{2} + 4 n - 4 - n + 2}{2} = \frac{n^{2} + 3 n - 2}{2}$

So, $α = \sum_{n = 1}^{10} {(\frac{n^{2} + 3 n - 2}{2})}^{2} and β = \sum_{n = 1}^{10} n^{4}$

$\Rightarrow 4 α = \sum_{n = 1}^{10} (n^{4} + 9 n^{2} + 4 + 6 n^{3} - 12 n - 4 n^{2})$ and $β = \sum_{n = 1}^{10} n^{4}$

Now, $4 α - β = \sum_{n = 1}^{10} (5 n^{2} + 6 n^{3} - 12 n + 4)$

$\Rightarrow 4 α - β = 5 \sum_{n = 1}^{10} n^{2} + 6 \sum_{n = 1}^{10} n^{3} - 12 \sum_{n = 1}^{10} n + 4 \times 10$

$= 5 \times \frac{10 (10 + 1) (20 + 1)}{6} + 6 {(\frac{10 \times 11}{2})}^{2} - 12 (\frac{10 \times 11}{2}) + 40$

$= 353 \times 55 + 40$

So, $k = 353$

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