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Ok so I'm trying to solve this mathematical induction problem and so far I got to a point but I can’t go any further the original problem is

Prove that for any positive integer n $$\sum_{j=1}^n j^3 = (\frac {n(n+1)} {2})^2$$

I done the problem up until i have

$$\sum_{j=1}^{k+1} j^3 = \sum_{j=1}^k j^3 + (k+1)^3 = (\frac {k(k+1)} {2})^2 + (k+1)^3$$

somehow i need to turn

$(\frac {k(k+1)} {2})^2 + (k+1)^3$

into

$(\frac {(k+1)(k+2)} {2})^2$

but I'm stuck on how to proceed

user3491432
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3 Answers3

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We have $(\frac{k(k+1)}{2})^2+(k+1)^3=(\frac{k(k+1)}{2})^2+\frac{4(k+1)^3}{4}=\frac{k^2(k+1)^2+4(k+1)^3}{4}=(k+1)^2 \cdot \frac{k^2+4(k+1)}{4}=(\frac{(k+1)(k+2)}{2})^2$

So you only need to work on your expression a little bit and you are done. That's very common and summation identies proved by induction often need algebraic manipulations at the end.

Arthr
  • 413
0

You are just there, the only thing you need to do is develop $(k+1)^3$ add it to the other term and simplify: $$ \frac{(k)^2(k+1)^2}{4}+(k+1)^3 = \frac{(k^4+2k^3+k^2) + 4(k^3+3k^2+3k+1)}{4} $$ Here things get a little messy, but if you add up all coefficients correctly you end up with: $$ \frac{k^4+6k^3+13k^2+12k+4}{4} $$

Now we develop the expression $(k+1)^2(k+2)^2 = (k^2+2k+1)(k^2+4k+2) = (k^4 +6k^3+13k^2+12k+4)$. From here, it is trivial to show that: $$ \frac{k^4+6k^3+13k^2+12k+4}{4} = \frac{(k+1)^2(k+2)^2}{4} $$ Just as we wanted.

A. Person
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So you got $[\frac{k(k+1)}{2}]^2 +[k+1]^3$

Taking $(k+1)^2$ common we get

\begin{align} &= (k+1)^2[\frac{k^2}{4} + (k+1)]\\ &= (k+1)^2[\frac{k^2+4k+4}{4}]\\ &= (k+1)^2[\frac{((k+1)+1)^2}{4}]\\ &= [\frac{(k+1)(k+2)}{2}]^2\\ \end{align}

Error_Akash
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