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Question

# Sum to n terms of the series 131+13+231+3+13+23+331+3+5+..... is

A
n24(n2+9n+14)
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B
n24(2n2+7n+15)
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C
n24(2n2+9n+13)
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D
n24(n2+11n+12)
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Solution

## The correct option is B n24(2n2+9n+13)We know,∑ni=1k=1+2+3+…+n=n(n+1)2 (sum of first n natural numbers)∑ni=1k2=12+22+32+…+n2=n(n+1)(2n+1)6 (sum of squares of the first n natural numbers)∑nk=1k3=13+23+33+…+n3=n2(n+1)24 (sum of cubes of the first n natural numbers). 1+3+.....+(2n−1)=∑nk=1(2k−1)=n2sum of the first odd n natural numbers). Let tr denote the rth term of the series, thentr=13+23+.....+r31+3+.....+(2r−1)=14r2(r+1)2r2=14(r+1)2⇒n∑r=1tr=14n∑r=1(r+1)2=14[n+1∑r=1r2+2r+1]=14[(n)(n+1)(2n+1)6+2n(n+1)2+n]=124[2n3+n2+2n2+n+6n2+6n+6n]=n24(2n2+9n+13)

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