The sum of the first 20 terms of the series 5 | vedclass

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(D) The given series is $5, 11, 19, 29, 41, \ldots$. Let the $n^{th}$ term be $T_n = an^2 + bn + c$. For $n=1, T_1 = a + b + c = 5$. For $n=2, T_2 = 4

Vedclass · Accepted Answer

$3520$

Vedclass · Answer

(D) The given series is $5, 11, 19, 29, 41, \ldots$. Let the $n^{th}$ term be $T_n = an^2 + bn + c$. For $n=1, T_1 = a + b + c = 5$. For $n=2, T_2 = 4a + 2b + c = 11$. For $n=3, T_3 = 9a + 3b + c = 19$. Subtracting equations: $(T_2 - T_1) = 3a + b = 6$ and $(T_3 - T_2) = 5a + b = 8$. Solving these,$2a = 2 \implies a = 1$,$b = 3$,and $c = 1$. Thus,$T_n = n^2 + 3n + 1$. The sum $S_{20} = \sum_{n=1}^{20} (n^2 + 3n + 1) = \sum_{n=1}^{20} n^2 + 3 \sum_{n=1}^{20} n + \sum_{n=1}^{20} 1$. Using formulas: $\sum n^2 = \frac{n(n+1)(2n+1)}{6}$,$\sum n = \frac{n(n+1)}{2}$. $S_{20} = \frac{20(21)(41)}{6} + 3 	imes \frac{20(21)}{2} + 20 = 2870 + 630 + 20 = 3520$.

The sum of the first $20$ terms of the series $5+11+19+29+41+\ldots$ is $..........$.

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