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Question Number 12107 by Joel576 last updated on 13/Apr/17

$${10}^3+{11}^3+{12}^3+...+{20}^3$$$$\mathrm{Is}\mathrm{there}\mathrm{any}\mathrm{ways}\mathrm{to}\mathrm{count}\mathrm{the}\mathrm{sum}$$$$\mathrm{of}\mathrm{that}\mathrm{sequence}\mathrm{without}\mathrm{sum}\mathrm{them}\mathrm{manually}?$$

Commented by FilupS last updated on 13/Apr/17

$$=2^3{5}^3+{11}^3+2^3{6}^3+{13}^3+2^3{7}^3+{15}^3$$$$+2^3{8}^3+{17}^3+2^3{9}^3+{19}^3+2^3{10}^3$$$$=2^3(5^3+6^3+7^3+8^3+9^3+{10}^3)$$$$+({11}^3+{13}^3+{15}^3+{17}^3+{19}^3)$$$$=2^3\underset{n=5}{\sum ^{10}}{n}^3+\underset{m=5}{\sum ^9}{(2m+1)}^3$$$$=2^3\underset{n=5}{\sum ^{10}}{n}^3+\underset{m=5}{\sum ^{10}}{(2m+1)}^3-21$$$$working$$

Commented by Joel576 last updated on 14/Apr/17

$$\mathrm{but}\mathrm{how}\mathrm{to}\mathrm{count}\left[\underset{m=5}{\sum ^{10}}{(2m+1)}^3\right]-{21}^3$$$$\mathrm{without}\mathrm{adding}\mathrm{them}\mathrm{manually}?$$

Answered by mrW1 last updated on 13/Apr/17

$$\underset{k=1}{\sum ^n}{k}^3=\frac{n^2{(n+1)}^2}{4}$$$${10}^3+{11}^3+{12}^3+...+{20}^3$$$$=\underset{k=1}{\sum ^{20}}{k}^3-\underset{k=1}{\sum ^9}{k}^3$$$$=\frac{{20}^2\times {21}^2-9^2\times {10}^2}{4}$$$$=\frac{{10}^2(2^2\times {21}^2-9^2)}{4}$$$$=25\times (2\times 21+9)\times (2\times 21-9)$$$$=25\times 51\times 33$$$$=42075$$

Commented by Joel576 last updated on 14/Apr/17

$$thankyouverymuch$$

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