If S n = 4 + 11 + 21 + 34 + 50 + . . . to n terms, then 1 60 ( S 29 - S 9 ) is equal to [2023]

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Solution

Q.
If $S_{n} = 4 + 11 + 21 + 34 + 50 + . . .$ to $n$ terms, then $\frac{1}{60} (S_{29} - S_{9})$ is equal to [2023]

1 223

2 220

3 226

4 227

Ans.
(1)

$Given, S_{n} = 4 + 11 + 21 + 34 + \dots up to n terms \dots (i)$

$Also, S_{n} = 4 + 11 + 21 + \dots + \dots up to n terms \dots (i i)$

$Subtracting (ii) from (i), we get:$

$0 = [4 + 7 + 10 + 13 + \dots + \dots] - a_{n}$

$∴$ $a_{n} = \frac{n}{2} [2 (4) + (n - 1) (3)] = \frac{n}{2} [3 n + 5] = \frac{3 n^{2} + 5 n}{2}$

$Now, S_{n} = \sum a_{n} = \frac{3}{2} \sum n^{2} + \frac{5}{2} \sum n$

$= \frac{3}{2} \frac{n \cdot (n + 1) (2 n + 1)}{6} + \frac{5}{2} \frac{n (n + 1)}{2} = \frac{n (n + 1)}{4} [(2 n + 1) + 5]$

$= \frac{n (n + 1) (2 n + 6)}{4} = \frac{n (n + 1) (n + 3)}{2}$

$Now, \frac{1}{60} [S_{29} - S_{9}] = \frac{1}{60} [\frac{(29) (30) (32)}{2} - \frac{(9) (10) (12)}{2}]$

$= \frac{1}{60} [13920 - 540] = \frac{13380}{60} = 223$

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