That's a lot of terms

Algebra Level 3

Determine the value of − 1 + 2 + 3 + 4 − 5 − 6 − 7 − 8 − 9 + . . . + 10000 -1 + 2 + 3 + 4 - 5 - 6 - 7 - 8 - 9 + ... + 10000 , where the signs change after each perfect square.


The answer is 1000000.

Sharky Kesa by Sharky Kesa , 19, United Kingdom

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

The sum can be express as:

S = ∑ n = 1 100 ( − 1 ) n ( ∑ k = 1 n 2 k − ∑ k = 1 ( n − 1 ) 2 k ) = ∑ n = 1 100 ( − 1 ) n ( n 2 ( n 2 + 1 ) 2 − ( n − 1 ) 2 ( ( n − 1 ) 2 + 1 ) 2 ) = ∑ n = 1 100 ( − 1 ) n ( 2 n 3 − 3 n 2 + 3 n − 1 ) = ∑ n = 1 100 ( − 1 ) n ( ( n − 1 ) 3 + n 3 ) = − 0 3 − 1 3 + 1 3 + 2 3 − 2 3 − 3 3 + ⋯ + 9 9 3 + 10 0 3 = 10 0 3 = 1000000 \begin{aligned} S & = \sum_{n=1}^{100} (-1)^n \left(\sum_{k=1}^{n^2} k - \sum_{k=1}^{(n-1)^2} k \right) \\ & = \sum_{n=1}^{100} (-1)^n \left(\frac {n^2(n^2+1)}2 - \frac {(n-1)^2((n-1)^2+1)}2 \right) \\ & = \sum_{n=1}^{100} (-1)^n \left(2n^3-3n^2+3n-1\right) \\ & = \sum_{n=1}^{100} (-1)^n \left((n-1)^3+n^3 \right) \\ & = -0^3-1^3+1^3+2^3-2^3-3^3+ \cdots +99^3+100^3 \\ & = 100^3 \\ & = \boxed{1000000} \end{aligned}

1 Helpful 0 Interesting 1 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...