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Question

Find the sum upto first 11 terms of the series 1.4.7 + 4.7.10 + 7.10.13 + . . . . . is___

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Solution

Let Tn be the nth term of given series.

Tn = (nth term of 1,4,7,.......) (nth term of 4,7,10, .......) (nth term of 7,10,13,.......)

= [1 + (n-1)3] [4 + 3(n-1)] [7 +(n-1)3]

= (3n - 2)(3n + 1) (3n + 4)

Let Vn = (3n - 2)(3n + 1) (3n + 4) (3n + 7)

Then Vn1 = (3n - 5)(3n - 2) (3n + 1) (3n + 4)

Vn - Vn1 = (3n - 2)(3n + 1) (3n + 4) [(3n + 7) - (3n - 5)]

=12Tn

Tn = 112 (Vn - Vn1)

Sn=Tn=nn=1112(VnVn1)

= 112(VnV0)

=112 ((3n1)(3n+2)(3n+4)(3n+7)+56)
So,sum of the first 11 terms

= 112 [(331)(33+2)(33+4)(33+7)+56]

= 138138


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