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

1+3+5+...+101

Question

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

1+3+5+...+101

zzr0ck3r · Answer

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

blackstreet23 · Answer

∑511 what does that mean?

anonymous · Answer

@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 $2(51)-1)=101$.
$$\sum_{n=1}^{51}(2n-1)=1+3+5+7+\cdots+99+101$$