What is the last number in this sum? 1+3+5+...+?=10,000

Question

Its confusing to me and I believed that I should have used the pascal triangles but it makes no sense to me now. Please help.

Answer 1

This is an arithmetic progression with first term a=1 and common difference d=2. If n is the number of terms, the last term is a+(n-1)d.

The sum S_n of the first n terms is a+a+d+a+2d+...+a+(n-1)d.

There are n/2 pairs of terms:

(a+a+(n-1)d)+(a+d+a+(n-2)d)+(a+2d+a+(n-3)d)+...

(2a+(n-1)d)+(2a+(n-1)d)+(2a+(n-1)d)+...

So each pair has the same sum and there are n/2 pairs, so:

S_n=(n/2)(2a+(n-1)d)=an+nd(n-1)/2.

But S_n=10000, and we know a=1 and d=2, so substitute for these:

n+2n(n-1)/2=n+n²-n=n²=10000, n=100.

The last number is a+(n-1)d=1+198=199,  making the series:
1+3+5+7+...+195+197+199.