Sum of products of consecutive odd numbers up to n terms

[avistein]

Find the sum upto n terms:

1.3.5+3.5.7+5.7.9....tn

I solve it this way:

tn=(2n-1)(2n+1)(2n+3)
Now can I take summation on both sides? How?

I mean when I add 2 on both sides the resultant is 0(2-2=0).Similarly the resultant summation will be zero?

And if I take summation I get one term as 3Ʃ.Now in a book I saw that it is 3n. Why? Summation of 3 will be 3 only as 3 is constant.Please explain.

I got this:


Ʃtn=Ʃ(2n-1)(2n+1)(2n+3)

Ʃtn=Ʃ[(4n^2-1)(2n+3)]
Ʃtn=Ʃ[8n^3 + 12n^2 - 2n - 3]

Ʃtn=Ʃ[8n^3] + Ʃ[12n^2] - Ʃ[2n] -Ʃ[3]

Ʃtn=8*Ʃ[n^3] + 12*Ʃ[n^2] - 2*Ʃ[n] - Ʃ[3]

Now do I require to write them as Ʃ[3] or 3Ʃ (putting a constant outside Ʃ).Please explain the whole summation process.I am stuck here.

 

[sweet springs]

Hi.

You are on the right way.

As for Ʃtn=8*Ʃ[n^3] + 12*Ʃ[n^2] - 2*Ʃ[n] - Ʃ[3], there are formula for Ʃ[n^3], Ʃ[n^2], Ʃ[n] and Ʃ[1].

 

[Stephen Tashi]

Summation of 3 will be 3 only as 3 is constant

No. Summing a constant depends on how many times you sum it.

For example,

$\sum_{i=1}^4 10 = 10 + 10 + 10 + 10 = (4)(10) = 40$