1).Find the sum of the n terms of the series Sn=1^2 + 3^2 +5^2+...+(2n-1)^2

Question

2).Find the sum to n terms of 1/(1.2.3) + 3/(2.3.4) + 5/(3.4.5) + 7/(4.5.6) +....

3) Sum to n terms,the series 1.3.5 + 2.4.6 + 3.5.7 +..

Answer 1

I will do the first one
We were lucky to have the general term, in most cases you have to find that first

Secondly you have to know some basic summation formulas
∑ k ,where k = 1 to n is n(n+1)/2
∑ k^2 , where k = 1 to n is n(n+1)(2n+1)/6
∑ c , where k = 1 to n is kc , where c was a constant

So we want ∑ (2n-1)^2
= ∑(4n^2 - 4n + 1)
= 4∑ n^2 - 4∑ n + ∑ 1
= 4n(n+1)(2n+1)/6 - 4n(n+1)/2 + n(1)
= (4n(n+1)(2n+1) - 12n(n+1) + 6n)/6
= ..
I will let you simplify this

Answer 2

Finding the sum of the first 2 terms of the series with the above formula won't get you the right answer

Answer 3

yes it will
Sum(1) = (4(2)(3) - 12(2) + 6)/6 = 1
sum(2) = (8(3)(5) - 24(3) + 12)/6 = 10
sum(3) = 12(4)(7) - 36(4) + 18)/6 = 35
etc.

It works! I am sure!