10-3-11-3-12-3-20-3-Is-there-any-ways-to-count-the-sum-of-that-sequence-without-sum-them-manually-

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

10^{3} + 11^{3} + 12^{3} + \dots + 20^{3}

Is there any ways to count the sum

of that sequence without sum them 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}} {(2 m + 1)}^{3}

= 2^{3} \underset{n = 5}{\sum^{10}} n^{3} + \underset{m = 5}{\sum^{10}} {(2 m + 1)}^{3} - 21

w o r k i n g

Commented by Joel576 last updated on 14/Apr/17

but how to count [\underset{m = 5}{\sum^{10}} {(2 m + 1)}^{3}] - 21^{3}

without adding them 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} + \dots + 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

t h a n k y o u v e r y m u c h

Facebook Tweet Pin