Question

The sum of the series 1.3.5+2.5.8+3.7.11+........Upto n terms is

Answer 1

Step-by-step explanation:

Appendix contains solution.

Answer 2

The sum of the series 1.3.5+2.5.8+3.7.11+... Upto n terms is 3 (n)²(n + 1)²/2  + 7(n)(n + 1)(2n+1)/6 + 2n(n + 1)

Solution:

1.3.5+2.5.8+3.7.11+...

It can be written as

∑ k(2k + 1)(3k + 2)   where k from 1 to n

On Expanding

= ∑ k(6k² + 7k + 4)

= ∑ (6k³ + 7k² + 4k)

= ∑6k³ + ∑7k² + ∑4k

= 6∑k³ + 7∑k² +  4∑k

∑k³ = (n)²(n + 1)²/4    where k from 1 to n

∑k² = (n)(n + 1)(2n+1)/6  where k from 1 to n

∑k = n(n + 1)/2   where k from 1 to n

= 6 (n)²(n + 1)²/4  + 7(n)(n + 1)(2n+1)/6 + 4n(n + 1)/2

= 3 (n)²(n + 1)²/2  + 7(n)(n + 1)(2n+1)/6 + 2n(n + 1)

Hence The sum of the series 1.3.5+2.5.8+3.7.11+... Upto n terms is 3 (n)²(n + 1)²/2  + 7(n)(n + 1)(2n+1)/6 + 2n(n + 1)