SOLUTION: Find the sum of the first n terms of the following series 3X1^2 X5X2^2+7X3^2+---

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Question 984368: Find the sum of the first n terms of the following series
3X1^2 X5X2^2+7X3^2+---
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
I believe what you really need to calculate is
S%28n%29=3%2A1%5E2%2B5%2A2%5E2%2B7%2A3%5E2%2B%22...%22%2B%282n%2B1%29%2An%5E2


Once you figure out that formula,
S%28n%29=%281%2F2%29n%5E4%2B%284%2F3%29n%5E3%2Bn%5E2%2B%281%2F6%29n can be proven by induction, like this:
1) The formula S%28n%29=%281%2F2%29n%5E4%2B%284%2F3%29n%5E3%2Bn%5E2%2B%281%2F6%29n is true for n=1 .
For n=1 , S%281%29=3%2A1%5E2=3
and the formula S%28n%29=%281%2F2%29n%5E4%2B%284%2F3%29n%5E3%2Bn%5E2%2B%281%2F6%29n ,
which tells us that
,
which makes the formula true tor n=1 .
2) We can prove that if S%28n%29=%281%2F2%29n%5E4%2B%284%2F3%29n%5E3%2Bn%5E2%2B%281%2F6%29n were true for any n=k whole number, it would be true for n=k%2B1 .
Term number k%2B1 is
.
If S%28k%29=%281%2F2%29k%5E4%2B%284%2F3%29k%5E3%2Bk%5E2%2B%281%2F6%29k , then

=
and that show that the formula is true for n=k%2B1 , because the formula says that

=
=
=
=%281%2F2%29k%5E4%2B%2810%2F3%29k%5E3%2B8k%5E2%2B%2849%2F6%29k%2B3

HOW WE CAN GET TO THAT FORMULA:
The sum of terms are of the form n%5Ep with p%3E=1 is a polynomial of degree p%2B1 in n with no independent term.
,
, and so on.
So, the sum of terms of the form 2n%2B1%29n%5E2=2n%5E3%2Bn%5E2 should be
S%28n%29=A%2An%5E4%2BB%2An%5E3%2BC%2An%5E2%2BD%2An
Knowing that
S%281%29=3%2A1%5E2=3%2A1=3 ,
S%282%29=3%2A1%5E2%2B5%2A2%5E2=3%2B5%2A4=3%2B20=23 ,
S%283%29=3%2A1%5E2%2B5%2A2%5E2%2B7%2A3%5E2=23%2B7%2A9=23%2B63=86 , and
S%284%29=3%2A1%5E2%2B5%2A2%5E2%2B7%2A3%5E2%2B9%2A4%5E2=86%2B9%2A16=86%2B144=230 ,
we can find A , B , C , and D by solving