Question

Find the sum to n terms of the series 1square +3square +5square +........to n terms

Answer 1

1^2+3^2+5^2+7^2+.....n^2

so.. first of all.. we know that the formula for the sum of square of n terms is n(n+1)(2n+1) /6

1^2+2^2+3^2+4^2+....(n-1)^2 + n^2 can be written as

1^2+3^2+5^2+..n^2 +(2^2+4^2+6^2+.... (n-1)^2)

now.. we need to substract the term in the parentheses from the sum of square of n terms in order to get our answer

if i take 2^2 common from the parentheses.. we get
2^2(1+2^2+3^2..... (n-1/2)^2)
that is..
n(n+1)(2n+1)/6 - 4[n-1/2(n+1/2)n/6]

n(n+1)/6[(2n+1) -n-1]=n^2(n+1)/6

there might be error in the simplification in the last step.. since i am not comfertable with phone... do check out by doing things in the exact same process- yourself..

hope you like it

Answer 2

Answer:

answer is the above pic.i couldn't crop with solution. please mark as brainliest