200 Followers Problem!

Algebra Level 3

1 2 + ( 1 2 + 2 2 ) + ( 1 2 + 2 2 + 3 2 ) + + ( 1 2 + + 19 9 2 + 20 0 2 ) = ? 1^2 + (1^2 + 2^2) + (1^2 + 2^2 + 3^2)+\ldots + (1^2 + \ldots + 199^2 + 200^2) = \ ?


The answer is 136016700.

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4 solutions

Ikkyu San
Jul 31, 2015

The above expression can be represented by n = 1 200 k = 1 n k 2 \displaystyle\sum_{n=1}^{200}\sum_{k=1}^nk^2 .

And k = 1 n k 2 = n ( n + 1 ) ( 2 n + 1 ) 6 \color{#624F41}{\displaystyle\sum_{k=1}^nk^2=\dfrac{n(n+1)(2n+1)}6} .

Thus,

n = 1 200 k = 1 n k 2 = n = 1 200 n ( n + 1 ) ( 2 n + 1 ) 6 = n = 1 200 2 n 3 + 3 n 2 + n 6 = n = 1 200 ( 1 3 n 3 + 1 2 n 2 + 1 6 n ) = 1 3 n = 1 200 n 3 + 1 2 n = 1 200 n 2 + 1 6 n = 1 200 n = 1 3 ( 200 ( 201 ) 2 ) 2 + 1 2 ( 200 ( 201 ) ( 401 ) 6 ) + 1 6 ( 200 ( 201 ) 2 ) = 1 3 ( 404010000 ) + 1 2 ( 2686700 ) + 1 6 ( 20100 ) = 134670000 + 1343350 + 3350 = 136016700 \begin{aligned}\displaystyle\sum_{n=1}^{200}\color{#624F41}{\sum_{k=1}^nk^2}=&\ \displaystyle\sum_{n=1}^{200}\color{#624F41}{\dfrac{n(n+1)(2n+1)}6}\\=&\ \displaystyle\sum_{n=1}^{200}\dfrac{2n^3+3n^2+n}6\\=&\ \displaystyle\sum_{n=1}^{200}\left(\dfrac13n^3+\dfrac12n^2+\dfrac16n\right)\\=&\ \dfrac13\color{#D61F06}{\displaystyle\sum_{n=1}^{200}n^3}+\dfrac12\color{#20A900}{\displaystyle\sum_{n=1}^{200}n^2}+\dfrac16\color{#3D99F6}{\displaystyle\sum_{n=1}^{200}n}\\=&\ \dfrac13\color{#D61F06}{\left(\dfrac{200(201)}2\right)^2}+\dfrac12\color{#20A900}{\left(\dfrac{200(201)(401)}6\right)}+\dfrac16\color{#3D99F6}{\left(\dfrac{200(201)}2\right)}\\=&\ \dfrac13\color{#D61F06}{(404010000)}+\dfrac12\color{#20A900}{(2686700)}+\dfrac16\color{#3D99F6}{(20100)}\\=&\ 134670000+1343350+3350=\boxed{136016700}\end{aligned}

are you a sigma lover?

Akash singh - 5 years, 7 months ago

Two equivalent ways to obtain the result:

E x p r e s s i o n = 200 1 2 + 199 2 2 + . . . + 1 20 0 2 , = i = 1 n ( 201 i ) i 2 = i = 1 n ( 210 i 2 i 3 ) = 201 n ( n + 1 ) ( 2 n + 1 ) 6 ( n ( n + 1 ) 2 ) 2 = n ( n + 1 ) 2 { 201 2 n + 1 3 n ( n + 1 ) 2 } B u t n = 200 , E x p r e s s i o n = 200 201 2 { 401 201 3 200 201 2 } = 136016700. Expression=200*1^2+ 199*2^2 + ... +1*200^2, \\ \displaystyle= \sum_{i=1}^{n} (201-i)*i^2=\sum_{i=1}^{n} (210*i^2-i^3) \\ =201*\dfrac{n*(n+1)*(2n+1)} 6 -\Big (\dfrac{n*(n+1)} 2 \Big )^2\\ =\dfrac{n*(n+1)} 2*\Big \{201*\dfrac{2n+1} 3 -\dfrac {n*(n+1)} 2 \Big \} \\ But~n=200,\\ \therefore~~ Expression= \dfrac{200*201} 2*\Big \{\dfrac{401*201} 3 -\dfrac {200*201} 2 \Big \} \\ =\Large~~~\color{#D61F06}{136 016 700}.

Johnny Lynch
Jul 30, 2015

The basics

Then figure it out. Sorry for the handmade

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