Deduction of Sigma

• MHB
• ChelseaL

ChelseaL

Deduce that
n
Sigma (2r)^3 = 2n^2 (n+1)^2
r=1

I'm not exactly sure how to start.

Deduce that
n
Sigma (2r)^3 = 2n^2 (n+1)^2
r=1

I'm not exactly sure how to start.

$$\displaystyle \sum_{r=1}^n\left((2r)^3\right)=8\sum_{r=1}^n\left(r^3\right)=\,?$$

The result of my previous thread was 1.

The result of my previous thread was 1.

https://mathhelpboards.com/discrete-mathematics-set-theory-logic-15/mathematical-induction-sigma-notation-23918.html

I assumed the two similarly named accounts from the same country were in some way related.

Oh right sorry, I got locked out of my previous account. The part with the k + 1 case?

Oh right sorry, I got locked out of my previous account. The part with the k + 1 case?

$$\displaystyle \sum_{k=1}^n\left(k^3\right)=\frac{n^2(n+1)^2}{4}$$

If we are allowed to use that formula, then this problem become very straightforward. :)

1
sigma r^3 = 1^2(1+1)^2/4?
r=1

1
sigma r^3 = 1^2(1+1)^2/4?
r=1

No, what I mean is:

$$\displaystyle \sum_{r=1}^n\left((2r)^3\right)=8\sum_{r=1}^n\left(r^3\right)=8\left(\frac{n^2(n+1)^2}{4}\right)=2n^2(n+1)^2$$

Deduce that
n
Sigma (2r)^3 = 2n^2 (n+1)^2
r=1

method 1

$\displaystyle 8 \sum_{r=1}^n r^3 = 8\color{red}{(1 + 2^3 + 3^3 + \, ... \, +n^3})$

sequence of partial sums for the series in parentheses is the triangular numbers squared ...

$1, 9, 36, 100, ... = 1^2, 3^2, 6^2, 10^2, ... , \left[\dfrac{n(n+1)}{2}\right]^2 = \dfrac{n^2(n+1)^2}{4}$

$\displaystyle 8 \sum_{r=1}^n r^3 = 8 \cdot \dfrac{n^2(n+1)^2}{4} = 2n^2(n+1)^2$

method 2 (induction) ...

$\displaystyle \sum_{r=1}^1 (2r)^3 = 2^3 = 8 = 2(1)^2 \cdot (1+1)^2$

assume true for $n$, show true for $(n+1)$

$\displaystyle \sum_{r=1}^{n+1} (2r)^3 = \sum_{r=1}^n (2r)^3 + [2(n+1)]^2 = 2n^2(n+1)^2 + [2(n+1)]^3 = 2(n+1)^2\bigg[n^2+4(n+1)\bigg] = 2(n+1)^2(n^2+4n+4) = 2(n+1)^2(n+2)^2$