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Question

The sum of 12.2213+22.3213+23+32.4213+23+33+.... upto n terms is equal to :

A
n−1n
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B
nn+1
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C
n+1n+2
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D
n+1n
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Solution

The correct option is B nn+1
12.2213+22.3213+23+32.4213+23+33+........+n2(n+12)13+23+.......+n3So,Tr=r2(r+12)13+23+.......+r3⇒14r(r+1)[r(r+1)2]2=r(r+1)4r(r+1)24Tr=r+1−rr(r+1)=1r−1r+1Sn=∑nr=1Tr=∑r=11r−1r+1=1−12+12−13+13−14+.....+1n−1n+1Sn=11−1n+1=n+1−1n+1=nn+1

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