-1

Show by induction that the following formulas hold for

$$ \sum_{i=1}^n i^3 = \frac{n^2(n+1)^2}{4} $$

Not sure how to go about this problem. Can someone help please? Thanks

user139388
  • 3,449
Alex
  • 31
  • http://math.stackexchange.com/questions/48080/proof-that-sum-limits-k-1nk2-fracnn12n16 – lab bhattacharjee Apr 06 '14 at 04:04
  • If someone can give me a head start, that will be great! – Alex Apr 06 '14 at 04:13
  • 2
    I don't mean to be rude, but this is your fourth question in a row concerning proofs using induction. Are you reading the answers to your previous questions ? If so, how come you still haven't got the slightest idea on how to proceed ? – Amateur Apr 06 '14 at 04:16

1 Answers1

0

The base case: $$\sum_{i=1}^1 i^3 = 1^3 = \frac{1^2 2^2}{4}.$$ Now suppose that the result holds for all $k \leq n$. Then \begin{align} \sum_{i=1}^{n+1} i^3 &= \sum_{i=1}^n i^3 + (n+1)^3 \\ &= \frac{n^2 (n+1)^2}{4} + (n+1)^3 \\ &= (n+1)^2 \left( \frac{n^2 }{4} + (n+1) \right) \\ &= \frac{(n+1)^2}{4}(n^2 + 4n + 4) \\ &= \frac{(n+1)^2}{4}(n+2)^2 \\ &= \frac{(n+1)^2(n+2)^2}{4}. \end{align}

user139388
  • 3,449
  • Got it. Thanks for the help! – Alex Apr 06 '14 at 04:25
  • No problem! Glad to help :) – user139388 Apr 06 '14 at 04:25
  • I got n²(n+1)² + 4(n+1)^3/4 so far. I would be my next step? Thanks – Alex Apr 06 '14 at 04:45
  • I've added one more line to bring you a bit closer to the solution (essentially, I factored $(n+1)^2/4$ out of the two terms in the last sum) – user139388 Apr 06 '14 at 04:47
  • I'm actually little confused of how to do it this way. using the above I have (n+1)²/4 (n+2)² so far. – Alex Apr 06 '14 at 05:08
  • Okay, I've edited my answer to include the full solution :) – user139388 Apr 06 '14 at 05:11
  • Thanks for the help. But how would (n+2)² can turn to n² at the end. Are we not suppose to end up with n²(n+1)²/4 instead of (n+1)²(n+2)²/4 ?? – Alex Apr 06 '14 at 05:19
  • No, because, as you'll notice, I'm summing up to $n+1$. I've assumed that the result holds for $n$, which is the term you reference, and want to prove it for $n+1$. So change your $n$ to $n+1$ and your $n+1$ to $n+2$. – user139388 Apr 06 '14 at 05:21