Question

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

Answer 1

Let $T_n$ be the nth term of given series.

$T_n$ = (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 $V_n$ = (3n - 2)(3n + 1) (3n + 4) (3n + 7)

Then $V_{n-1}$ = (3n - 5)(3n - 2) (3n + 1) (3n + 4)

$V_n$ - $V_{n-1}$ = (3n - 2)(3n + 1) (3n + 4) [(3n + 7) - (3n - 5)]

=12$T_n$

$\Rightarrow$ $T_n$ = $\frac{1}{12}$ ($V_n$ - $V_{n-1}$)

$S_n=\sum T_n={\sum }_{n=1}^n\frac{1}{12}(V_n-V_{n-1})$

= $\frac{1}{12}(V_n-V_0)$

=$\frac{1}{12}$ $\left(\right(3n-1\left)\right(3n+2\left)\right(3n+4\left)\right(3n+7)+56)$
So,sum of the first 11 terms

= $\frac{1}{12}$ $\left[\right(33-1\left)\right(33+2\left)\right(33+4\left)\right(33+7)+56]$

= 138138