Prove series to the power of 5 by induction

Question

Prove by induction:

$$ 1^5 + 2^5 + 3^5 + . . . + n^5 = \frac{n^2 (n + 1)^2 (2n^2 + 2n − 1)}{12} $$

I tried to solve it to some extent. Can anyone help me out please?

Answer 1

Hint: For the inductive step, you want to prove that if:

$$ 1^5 + 2^5 + 3^5 + . . . + k^5 = \frac{k^2 (k + 1)^2 (2k^2 + 2k − 1)}{12} \tag{1}$$

Then the following is also true:

$$ 1^5 + 2^5 + 3^5 + . . . + k^5+(k+1)^5 = \frac{(k+1)^2 (k + 2)^2 (2k^2 + 6k +3)}{12} \tag{2}$$

So let us first assume that equation $(1)$ is true. then:

$$ 1^5 + 2^5 + 3^5 + . . . + k^5+(k+1)^5 = \frac{k^2 (k + 1)^2 (2k^2 + 2k − 1)}{12}+(k+1)^5 $$ $$ 1^5 + 2^5 + 3^5 + . . . + k^5+(k+1)^5 = \frac{k^2 (k + 1)^2 (2k^2 + 2k − 1)+12(k+1)^5}{12}$$ $$ 1^5 + 2^5 + 3^5 + . . . + k^5+(k+1)^5 = \frac{(k+1)^2(k^2(2k^2+2k-1)+12(k+1)^3)}{12}$$

I will leave you to finish off the proof. The only thing you need to do now is to prove $k^2(2k^2+2k-1)+12(k+1)^3=(k+2)^2(2k^2 + 6k +3)$.

Answer 2

I guess you might have verified it for $n=1$

Now assume it is true for $n$ and then try to prove for $n+1$. Break the summation into two parts. One having $n$ terms and the other part being the $n+1^{th}$ term itself. Substitute the sum of $n$ terms from your assumption.

According to me, you should not have much difficulty in solving the problem by induction. There are some other elegant methods too as linked below.

1.Method of undetermined coefficients

2.Faulhaber's formula and other generalizations

Answer 3

$k^6-(k-1)^6=6k^5-15k^4+20k^3-15k^2+6k-1$, therefore, $$ n^6=6\sum_{k=1}^{n}k^5-15\sum_{k=1}^{n}k^4+20\sum_{k=1}^{n}k^3-15\sum_{k=1}^{n}k^2+6\sum_{k=1}^{n}k-n $$ I assume you know what $\sum_{k=1}^{n}k^4$, $\sum_{k=1}^{n}k^3$,$\sum_{k=1}^{n}k^2$ are. Actually you can apply this method again. Although it's not induction, it's much more convenient.