Double summation

Number Theory Level 3

Find the value of

∑ k = 1 99 ( ∑ i = 1 k i ) \sum _{ k=1 }^{ 99 }{ (\sum _{ i=1 }^{ k }{ i } ) }


The answer is 166650.

K. J. W. by K. J. W. , 20, Singapore

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3 solutions

Ikkyu San
Jul 13, 2015

∑ k = 1 99 ∑ i = 1 k i = ∑ k = 1 99 k ( k + 1 ) 2 = ∑ k = 1 99 k 2 + k 2 = 1 2 ∑ k = 1 99 ( k 2 + k ) = 1 2 ( ∑ k = 1 99 k 2 + ∑ k = 1 99 k ) = 1 2 ( ( 99 ) ( 100 ) ( 199 ) 6 + ( 99 ) ( 100 ) 2 ) = 1 2 ( 328350 + 4950 ) = 1 2 ( 333300 ) = 166650 \begin{aligned}\displaystyle\sum_{k=1}^{99}\sum_{i=1}^{k}i=&\ \displaystyle\sum_{k=1}^{99}\dfrac{k(k+1)}2\\=&\ \displaystyle\sum_{k=1}^{99}\dfrac{k^2+k}2\\=&\ \dfrac12\displaystyle\sum_{k=1}^{99}(k^2+k)\\=&\ \dfrac12\left(\displaystyle\sum_{k=1}^{99}k^2+\displaystyle\sum_{k=1}^{99}k\right)\\=&\ \dfrac12\left(\dfrac{(99)(100)(199)}6+\dfrac{(99)(100)}2\right)\\=&\ \dfrac12(328350+4950)\\=&\ \dfrac12(333300)=\boxed{166650}\end{aligned}

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Summation upto n terms = n(n+1)/2. Summation of summations=(1^2+2^2+. . .+99^2+1+2+. . +99)/2. Summation of squares upto n terms=n(n+1)(2n+1)/6. Use these.

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Was my solution too poor to earn an upvote??

Chandrachur Banerjee - 6 years, 9 months ago
K. J. W.
Jul 16, 2014

Apologies for the typo in the original question.

The sum is

+1

+1+2

+1+2+3

+1+2+3+4

+1+2+3+4+5

...

+1+2+...+99

=1x99+2x98+3x97+...+98x2+99x1

=2(1x99+2x98+..+49x51)+50x50

=2((50+1)(50-1)+(50+2)(50-2)+...+(50+49)(50-49))+50x50

=2(49(50^2)-(1^2+2^2+...+49^2))+50x50

=2(122500-((49)(50)(99))/6)+50x50

=2(122500-40425)+50x50

=2(82075)+50x50

=164150+2500

=166650

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