Question

Find the sum of 'n' terms of the series

Answer 1

Heya User,

--> 1² + 4² + ... + upto nth term
So, we find the general term as:->
===> a = 1; d = 3; .'. nth Term = a+(n-1)d = [3n - 2]

Hence, we define the nth term as [ 3n - 2 ]...
--> ∑[3n-2]² = $\frac{[3n-2][3n-1][6n-3]}{6}$ <-- Ans.. (i)

---> Similarly, nth term of the second A.P. = (3n-1)(3n+2)(3n+5)
--> nth term of the second A.P. = (3n-1)(3n+2)(3n+5) 
===== [ 27n³ + 54n² + 9n - 10 ]

=>∑[ 27n³ + 54n² + 9n - 10 ] = 27∑n³ + 54∑n² + 9∑n - ∑10
========= $27 { \frac{n[n+1]} {4}}^{2} + 54 \frac{n[n+1][2n+1]}{6} + 9 \frac{n[n+1]}{2} - 10n$

This after solving becomes:-> 
----> $\frac{1}{4} n (3n+7)(9 n^{2} + 21n + 2)$ <--- Ans.. (ii)