Suppose the nth term of a series is $1+\frac{n}{2}+\frac{n^2}{2}$. If there are 20 terms in the series, then the sum of the series is equal to
1360
1450
1500
1560
Answer is Right!
Answer is Wrong!
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UPSC CAPF – 2021
$S_{20} = \sum_{n=1}^{20} (1+\frac{n}{2}+\frac{n^2}{2}) = \sum_{n=1}^{20} 1 + \frac{1}{2}\sum_{n=1}^{20} n + \frac{1}{2}\sum_{n=1}^{20} n^2$
Using standard summation formulas:
$\sum_{n=1}^{N} c = Nc$
$\sum_{n=1}^{N} n = \frac{N(N+1)}{2}$
$\sum_{n=1}^{N} n^2 = \frac{N(N+1)(2N+1)}{6}$
Here, N = 20.
$\sum_{n=1}^{20} 1 = 20 \times 1 = 20$
$\sum_{n=1}^{20} n = \frac{20(20+1)}{2} = \frac{20 \times 21}{2} = 10 \times 21 = 210$
$\sum_{n=1}^{20} n^2 = \frac{20(20+1)(2 \times 20 + 1)}{6} = \frac{20 \times 21 \times 41}{6} = \frac{10 \times 7 \times 41}{1} = 2870$
Now substitute these values back into the sum expression:
$S_{20} = 20 + \frac{1}{2}(210) + \frac{1}{2}(2870)$
$S_{20} = 20 + 105 + 1435$
$S_{20} = 125 + 1435 = 1560$.