11^3 + 12^3 + dots + 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 = \left[ \frac{k(k+1

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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 $\sum_{n=1}^{k} n^3 = \left[ \frac{k(k+1)}{2} \right]^2$: Sum $= \left[ \frac{20(21)}{2} \right]^2 - \left[ \frac{10(11)}{2} \right]^2$ $= (210)^2 - (55)^2$ $= 44100 - 3025 = 41075$. Since the last digit is $5$,the number $41075$ is divisible by $5$. Since the last digit is $5$,it is an odd integer. Therefore,the sum is an odd integer divisible by $5$.

$11^3 + 12^3 + \dots + 20^3$

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