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Find the sum 1^2+(1^2+2^2)+(1^2+2^2+3^2)...

Find the sum `1^2+(1^2+2^2)+(1^2+2^2+3^2)+` up to 22nd term.

Text Solution

Verified by Experts

The correct Answer is:
23276
`T_(n)=1^(2)+2^(2)+3^(2)+…+n^(2)`
`=(n(n+1)(2n+1))/6`
`=n/6(2n^(2)+3n+1)`
`=(2n^(3)+3n^(2)+n)/6`
`S_(n)=sumT_(n)`
`=1/6[2sumn^(3)+3sumn^(2)+sumn]`
`=1/6{2[(n(n+1))/2]^(2)+3xx(n(n+1)(2n+1))/6+(n(n+1))/2}`
`=1/6[(n^(2)(n+1)^(2))/2+(n(n+1)(2n+1))/6+(n(n+1))/2]`
`=1/6xx1/2n(n+1)[n(n+1)[n(n+1)+(2n+1)+1]`
`=(n(n+1))/12[n^(2)+3n+2]`
For n= 22
`S_(22)=(22xx23)/12[22^(2)+66+2]`
`=(22xx23)12[552]`
`=22xx23xx46`
=23276
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