Question

Use the formula S=n^2 to find each sum

1+3+5+...+101

Answer 1

\(\sum_1^{51}2n-1=2\sum_1^{51}n-\sum_1^{51}1=2(\frac{n(n+1)}{2})-51=2*51*26-51= 2601\)

Answer 2

∑511 what does that mean?

Answer 3

@blackstreet23 you're summing up the odd natural numbers between and including 1 to 101.

\(2n-1\) is an odd number for integer \(n\). Plug in \(n=1,2,3,...\) and you have \(2n-1=1,3,5,...\).

The \(\sum\) symbol is shorthand for summations like this one. For example,
\[\sum_{n=1}^4n=1+2+3+4\]
Similarly,
\[\sum_{n=1}^4(2n-1)=1+3+5+7\]
The sum in question starts with \(n=1\) and ends with \(n=51\), since \(2(1)-1=1\) and \(2(51)-1)=101\).
\[\sum_{n=1}^{51}(2n-1)=1+3+5+7+\cdots+99+101\]