11^3 + 12^3 + .... + 20^3 | vedclass

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(B) The sum is given by $\sum_{n=11}^{20} n^3 = \sum_{n=1}^{20} n^3 - \sum_{n=1}^{10} n^3$.Using the formula $\sum_{n=1}^{k} n^3 = [\frac{k(k+1)}{2}]^

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Is an odd integer divisible by $5$

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(B) The sum is given by $\sum_{n=11}^{20} n^3 = \sum_{n=1}^{20} n^3 - \sum_{n=1}^{10} n^3$.Using the formula $\sum_{n=1}^{k} n^3 = [\frac{k(k+1)}{2}]^2$:For $k=20$,$\sum_{n=1}^{20} n^3 = [\frac{20(21)}{2}]^2 = (210)^2 = 44100$.For $k=10$,$\sum_{n=1}^{10} n^3 = [\frac{10(11)}{2}]^2 = (55)^2 = 3025$.Therefore,the sum is $44100 - 3025 = 41075$.Since the number $41075$ ends in $5$,it is an odd integer and is divisible by $5$.

$11^3 + 12^3 + .... + 20^3$

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