Find the sum till infinity 1*3*5 + 3*5*7 + 5*7*9...?

1 Answer

Sum of the series tends to oo.

Explanation:

It is evident that sum of the series tends to oo. However, let us find sum up to n terms.

n^(th) term of the series 1*3*5+3*5*7+5*7*9+.... is

(2n-1)(2n+1)(2n+3) and we have to find sum_1^n((2n-1)(2n+1)(2n+3))

As (2n-1)(2n+1)(2n+3)

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

= 8n^3+12n^2-2n-3

Hence sum_1^n(8n^3+12n^2-2n-3)

= 8sum_1^n n^3+12sum_1^n n^2-2sum_1^n n-3n

= 8(n^2(n+1)^2)/4+12(n(n+1)(2n+1))/6-2(n(n+1))/2-3n

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

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

= 2n^4+4n^3+2n^2+4n^3+2n^2+4n^2+2n-n^2-n-3n

= 2n^4+8n^3+7n^2-2n

It is apparent that as n->oo sum of the series tends to oo.