Question

How do i find a formula for "1+3+5+7+......+(2n-1)

Answer 1

Explanation:

common difference d=3-1=2

first term a=1

an=a+(n-1)d

2n-1=1+(l-1)2

2n-1=1+2l-2

2n-1=2l-1

l=n

(i used l for number of terms)

number of terms=n

`S_(n)=(n)/(2) (first ~term+last~term)\\=(n)/(2) (1+2n-1)\\=n^2`