# Sum of cubes.

• Jul 27th 2010, 01:19 PM
fxs12
Sum of cubes.
$\displaystyle \begin{array}{l} prove\;that\; \\ \\ 1^3 + 2^3 + 3^3 + ............ + 100^3 = (\frac{{100 \times 101}}{2})^2 \\ \\ \end{array}$
• Jul 27th 2010, 01:22 PM
mr fantastic
Quote:

Originally Posted by fxs12
$\displaystyle \begin{array}{l} prove\;that\; \\ \\ 1^3 + 2^3 + 3^3 + ............ + 100^3 = (\frac{{100 \times 101}}{2})^2 \\ \\ \end{array}$

Use Google to fnd the sum of cubes of formula. Now apply that formula.
• Jul 27th 2010, 01:43 PM
Also sprach Zarathustra
1^3+2^3+3^3+...+n^3=(1^2+2^2+3^2+...+n^2)^2={(n(n+ 1)/2}^2

Someway to prove the formula:

An^4+Bn^3+Cn^2+Dn+E=S_n

put:
n=1,2,3,4

and solve the system of 4 equations, find A,B,C,D,E.
• Jul 27th 2010, 01:52 PM
Soroban
Hello, fxs12!

How about an inductive proof?

Quote:

$\displaystyle \text{Prove that: }\;1^3 + 2^3 + 3^3 + \hdots + 100^3 \:=\: \left(\dfrac{100 \times 101}{2}\right)^2$

$\displaystyle \text{We want to prove that: }\;1^3 + 2^3 + 3^3 + \hdots + n^3 \:=\:\left[\dfrac{n(n+1)}{2}\right]^2$

$\displaystyle \text{Establish }S(1):\;\;1^2 \:=\:\left[\frac{(1)(2)}{2}\right]^2$ . . . True.

$\displaystyle \text{Assume }S(k)\text{ is true: }\;1^3 + 2^3 + 3^3 + \hdots + k^3 \;=\;\left[\dfrac{k(k+1)}{2}\right]^2$

$\displaystyle \text{Add }(k+1)^3\text{ to both sides:}$

. . $\displaystyle 1^3 + 2^3 + 3^3 + \hdots + k^3 + (k+1)^3 \;=\;\left[\dfrac{k(k+1)}{2}\right]^2 + (k+1)^3$ .[1]

$\displaystyle \text{The right side is: }\:\dfrac{k^2(k+1)^2}{4} + (k+1)^3 \;=\;(k+1)^2\bigg[\dfrac{k^2}{4} + k + 1\bigg]$

. . . . . . . . . . . . $\displaystyle =\;(k+1)^2\bigg[\dfrac{k^2 + 4k + 4}{4}\bigg] \;=\;\dfrac{(k+1)^2(k+2)^2}{4}$

Therefore, [1] becomes: .$\displaystyle 1^3 + 2^3 + 3^3 + \hdots + (k+1)^3 \;=\;\left[\dfrac{(k+1)(k+2)}{2}\right]^2$

The inductive proof is complete.