500 again

Algebra Level 4

$1+ 4+10+ 20+35+ 56+ \ldots$ .

Find the sum of the above series up to $500^{th}$ term.

The answer is 2635531375.

1 solution

Maggie Miller
Jul 18, 2015

Call the terms in the series $a_1,a_2,\ldots$ . Note the difference $a_n-a_{n-1}$ is the $n$ -th triangular number, and $a_1$ is the first triangular number. Therefore, when summing the first $500$ terms, we are summing $501-n$ copies of the $n$ -th triangular number as $n$ ranges from $1$ to $500$ . Then the answer is $\sum_{k=1}^{500}\sum_{n=1}^{k}\frac{n(n+1)}{2}=\sum_{k=1}^{500}\frac{k(k+1)(k+2)}{6}=\frac{500\cdot 501\cdot 502\cdot 503}{24},$

i.e. the answer is $\boxed{2635531375}.$

great solution

Sai Ram - 5 years, 11 months ago

I think an extra step may make it more clear. The sum from n = 1 to n = k of n(n + 1)/2 could be written as (1/2) sum of first k squares + (1/2) sum of first k numbers = (1/2) [k(k + 1)(2k +1)/]6 + (1/2) [k(k+ 1)]/2 = (k^2 + k)*(2k + 1)/12 + (k^2 + k)/4 = [(k^2 + k)(2k + 1) + 3(k^2 + k)]/12 = (k^2 + k)(2k + 4)/12 = k(k + 1)(k + 2)/6. Ed Gray

Edwin Gray - 2 years, 4 months ago

I got it right by guessing. =)

Garv Khurana - 2 years, 4 months ago

500 again

The answer is 2635531375.

1 solution

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