Question

the sum of series 31+ 33+...+53

Answer 1

The series 31+33...+53  can be expressed as sigma notation ∑n=1103n∑n=1103n . This expression is read as the sum of 3n3n as nngoes from 11 to 10

Answer 2

Answer:

$31+33+...+53 = 504$

Step-by-step explanation:

Given series 31+33+...+53

is an A.P

First term (a) = 31

$common \: difference \:(d)=a_{2}-a_{1}\\=33-31\\=2$

$n^{th} \:term =a_{n}=53$

$\implies a+(n-1)d=53$

$\implies 31+(n-1)2=53$

$\implies (n-1)2=53-31$

$\implies (n-1)2=22$

$\implies (n-1)=\frac{22}{2}$

$\implies n= 11+1$

$\implies n = 12$

$Now,\\Sum \: of \: n \: terms (S_{n})= \frac{n}{2} (a+a_{n})$

$\implies S_{12}=\frac{12}{2}(31+53)\\=6\times 84\\=504$

Therefore,.

$31+33+...+53 = 504$

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