The sum of the first 65 odd, positive integers is?

1 answer:

7 0

$1+3+5+7+9+11+...\\\\This\ is\ an\ atihmetic\ progression\ where\ a_1=1\ and\ d=2.\\\\The\ sum:S_n=\frac{2a_1+(n-1)d}{2}\cdot n\\\\S_{65}=\frac{2\cdot1+(65-1)\cdot2}{2}\cdot65=\frac{2+64\cdot2}{2}\cdot65=\frac{2+128}{2}\cdot65\\\\=\frac{130}{2}\cdot65=65\cdot65=4225$

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