efekwulsemmay
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Nevermind, got the answer :D
I need to prove using mathematical induction that:
[tex]\sum^{n}_{i=1} i^{3}= 1^{3}+2^{3}+3^{3}+...+n^{3}=\left[\frac{(n+1)\cdot n}{2}\right]^{2}[/tex]
[tex]\sum^{n}_{i=1} i^{3}= 1^{3}+2^{3}+3^{3}+...+n^{3}=\left[\frac{(n+1)\cdot n}{2}\right]^{2}[/tex]
Ok, I have proved that the statement is true when n=1 and have gone on to assume that it must be true for any arbitrary value of n. I chose n=p. Thus:
[tex]\sum^{p}_{i=1} i^{3}= 1^{3}+2^{3}+3^{3}+...+p ^{3}=\left[\frac{(p+1)\cdot p}{2}\right]^{2}[/tex]
Where i am stuck is that I know i need to show that if the statement is true when n=p then it must be true for n=p+1. I have gotten as far as:
[tex]\sum^{p}_{i=1} i^{3}= 1^{3}+2^{3}+3^{3}+...+p^{3}+(p+1)^{3}=\left[\frac{(p+1)\cdot p}{2}\right]^{2}+(p+1)^{3}[/tex]
Homework Statement
I need to prove using mathematical induction that:
[tex]\sum^{n}_{i=1} i^{3}= 1^{3}+2^{3}+3^{3}+...+n^{3}=\left[\frac{(n+1)\cdot n}{2}\right]^{2}[/tex]
Homework Equations
[tex]\sum^{n}_{i=1} i^{3}= 1^{3}+2^{3}+3^{3}+...+n^{3}=\left[\frac{(n+1)\cdot n}{2}\right]^{2}[/tex]
The Attempt at a Solution
Ok, I have proved that the statement is true when n=1 and have gone on to assume that it must be true for any arbitrary value of n. I chose n=p. Thus:
[tex]\sum^{p}_{i=1} i^{3}= 1^{3}+2^{3}+3^{3}+...+p ^{3}=\left[\frac{(p+1)\cdot p}{2}\right]^{2}[/tex]
Where i am stuck is that I know i need to show that if the statement is true when n=p then it must be true for n=p+1. I have gotten as far as:
[tex]\sum^{p}_{i=1} i^{3}= 1^{3}+2^{3}+3^{3}+...+p^{3}+(p+1)^{3}=\left[\frac{(p+1)\cdot p}{2}\right]^{2}+(p+1)^{3}[/tex]
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