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Question

Find the sum of series to n terms:
125+236+347+.....

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Solution

125+236+347+.....

In each term of sequence, first, second and third number forms an A.P. with common difference 1,1 and 1 respectively.

Let Tn be the nth term of the given series, then
Tn={nth term of 1,2,3,...}×{nth term of 2,3,4,...}×{nth term of 5,6,7,...}

={1+(n1)×1}×{2+(n1)×1}×{5+(n1)×1}

=n(n+1)(n+4)

=(n2+n)(n+4)

=n3+4n2+n2+4n

Tn=n3+5n2+4n (1)

Now, let Sn ne the sum of n terms of the given series

We have:
Sn=Tn

=(n3+5n2+4n) [Using equation (1)]

=n3+5n2+4n

=n2(n+1)24+5n(n+1)(2n+1)6+4n(n+1)2

=n(n+1){n(n+1)4+56(2n+1)+2}

=n(n+1){3n(n+1)+2×5(2n+1)+2×1212}

=n(n+1){3n2+3n+20n+10+2412}

=n12(n+1)(3n2+23n+34)

Sum of series Sn=n12(n+1)(3n2+23n+34)


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