1. ## Sequences

Evaluate 1 x 2^2 + 3 x 4^2 + 5 x 6^2 +... to n terms
I have been using (2n-1)(2n^2) as my nth term. After finding my sum to n terms, i have been substituting a value in, continuously it is not coming out as intended. Possibly my nth term is incorrect. I have been using the rth term expressed as a polynomial in r method.
Thanks.

2. Originally Posted by Webby
Evaluate 1 x 2^2 + 3 x 4^2 + 5 x 6^2 +... to n terms
I have been using (2n-1)(2n^2) as my nth term. After finding my sum to n terms, i have been substituting a value in, continuously it is not coming out as intended. Possibly my nth term is incorrect. I have been using the rth term expressed as a polynomial in r method.
Thanks.
Your T_n will be = (2n-1)(4n^2)

$T_n = (2n-1)(2n)^2$

$S_n = \sum_{r=1}^x{(8n^3-4n^2)}$

I don't think you need help from here, ask it if you feel so

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EDIT:

$S_n = \sum_{r=1}^x{(8n^3-4n^2)}$

$= \frac{8(x(x+1))^2}{4} - 4 \frac{(x(x+1)(2x+1))}{6}$

$= 2(x(x+1))^2 - 2 \frac{(x(x+1)(2x+1))}{3}$

Thus only simplification is left

3. The sum of 4n^2, would be 4n(n+1)(2n+1)/6 if i am not mistaken.
However i am coming up with a sum to n terms of 2n(n+1)(3n^2+n-1)/3. When substituting values n>1. I am coming out 4 ahead of what i should be. Not sure where i am going wrong.

4. Originally Posted by Webby
The sum of 4n^2, would be 4n(n+1)(2n+1)/6 if i am not mistaken.
However i am coming up with a sum to n terms of 2n(n+1)(3n^2+n-1)/3. When substituting values n>1. I am coming out 4 ahead of what i should be.

I had an idiotic typo , I have corrected it now

As far as answer is concerned

Sum of x terms $= 2(x(x+1))^2 - 2 \frac{(x(x+1)(2x+1))}{3}$

Put 2 , sum should come out to be = 4+48 =52

So lets put x= 2 in the answer

$2(2(3))^2 - 2 \frac{(2(3)(5))}{3}$

$= 2(4*9) - 2 \frac{30}{3}$

$= 72 - 20$

$= 52$
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Your mistake was that you considered the nth term

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

However it should be = (2n-1)(2n)^2

5. Hello, Webby!

Evaluate: . $1\cdot2^2 + 3\cdot4^2 + 5\cdot6^2 + \hdots$ .to $n$ terms.

$\text{I have been using }(2n-1)\underbrace{(2n^2)}_{{\color{red}\text{no}}}\tex t{ as my }n^{th}\text{ term.}$

As you suspected, your $n^{th}$ term is wrong.
We want the square of ${\color{red}2}n\!:\;\;(2n)^2 = 4n^2$

ARDASH's work is excellent and correct!

To simplify that formula: . $S_n\;=\;2n^2(n+1)^2 - \tfrac{2}{3}n(n+1)(2n+1)$

. . Factor! . $\frac{2}{3}n(n+1)\bigg[3n(n+1) - (2n+1)\bigg]$

Therefore: . $S_n \;=\;\frac{2}{3}n(n+1)(3n^2+n-1)$