The sum sum_{k=1}^{20}(1+2+3+ldots+k) is | vedclass

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(C) The sum of the first $k$ natural numbers is given by the formula $\frac{k(k+1)}{2}$.We need to calculate $\sum_{k=1}^{20} \frac{k(k+1)}{2}$.This c

Vedclass · Accepted Answer

$1540$

Vedclass · Answer

(C) The sum of the first $k$ natural numbers is given by the formula $\frac{k(k+1)}{2}$.We need to calculate $\sum_{k=1}^{20} \frac{k(k+1)}{2}$.This can be written as $\frac{1}{2} \sum_{k=1}^{20} (k^2 + k)$.Using the summation formulas $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$ for $n=20$:$\sum_{k=1}^{20} k^2 = \frac{20(21)(41)}{6} = 2870$.$\sum_{k=1}^{20} k = \frac{20(21)}{2} = 210$.Therefore,the sum is $\frac{1}{2} (2870 + 210) = \frac{1}{2} (3080) = 1540$.

The sum $\sum_{k=1}^{20}(1+2+3+\ldots+k)$ is

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