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Algebra Level 2

For what value of n does z = 0 n z = 2016 ? \text{For what value of n does }\displaystyle{\sum_{z=0}^{n}{z}} \, = \, 2016 \, ?


The answer is 63.

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2 solutions

Rishabh Jain
Feb 1, 2016

z = 0 n z = n ( n + 1 ) 2 \Large\displaystyle\sum_{z=0}^{n}z=\dfrac{n(n+1)}{2} n ( n + 1 ) = 4032 = 63 × 64 \Large \therefore \color{#EC7300}{n}(\color{darkviolet}{n+1})=4032=\color{#EC7300}{63}\times \color{darkviolet}{64} Hence n = 63 \large\boxed{\color{#D61F06}{n=63}}

Sravanth C.
Feb 1, 2016

We know that sum of n n natural numbers is = z = 0 n z = n ( n + 1 ) 2 =\displaystyle\sum_{z=0}^{n}z=\dfrac{n(n+1)}{2} .

In this case,

z = 0 n z = n ( n + 1 ) 2 = 2016 n 2 + n 4032 = 0 n = 1 ± 1 ( 4 × 1 × 4032 2 = 1 ± 127 2 \displaystyle\sum_{z=0}^{n}z=\dfrac{n(n+1)}{2}=2016\\\implies n^2+n-4032=0\\ n=\dfrac{-1\pm\sqrt{1-(4\times 1\times 4032}}{2}=\dfrac{-1\pm127}2

Therefore, n = 63 \boxed{n=63} or n = 64 n=-64 , as n n can't be negative, the positive value is correct.

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