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Prove the sum of the cube of the first n integers(i.e. summation n^3) by mathematical induction

Posted February 14th, 2010 by Muideen
in

Prove by mathematical induction that

$$\sum_{r=1}^{n}r^3=\frac{n^2(n+1)^2}{4}$$
‹ Sigma notation Inequality ›

Comments

Proof:

On February 21st, 2010 alpha says:

Let P(n): $ 1^3+2^3+3^3.........................+n^3=\frac{n^2(n+1)^2}{4} $
STEP 1]P(1) is true.
STEP2]INDUCTION ASSUMPTION:
Let P(k) be true.
Then,P(k)=$ 1^3+2^3+3^3.........................+k^3=\frac{k^2(k+1)^2}{4} $
STEP3]To prove that P(k+1) is true.
i.e P(k+1)=$ 1^3+2^3+3^3......................+k^3+(k+1)^3=\frac{(k+1)^2(k+2)^2}{4} $
LHS= $ \frac{k^2(k+1)^2}{4}+(k+1)^3 $[from induction assumption $ 1^3+2^3+3^3.........................+k^3=\frac{k^2(k+1)^2}{4} $]
=$ \frac{(k+1)^2}{4}[k^2+4k+4] $
=$ \frac{(k+1)^2 (k+2)^2}{4} $
Hence, proved.

Hope I helped you!

-------------------x----------------------------
``Old theorems never die; they turn into definitions.''
----E. Hewitt

''ALPHA" =)

answer

On February 15th, 2010 Structure says:

see Sum of cubes

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