Prove that for any $n \in \mathbb{N}$, $\sum_{k=1}^n k^2(k+1)=\dfrac{n(n+1)(n+2)(3n+1)}{12}$

Question

I have written the LHS using the sum of arithmetic sequence formula and got n((2+n^2+n^3)/2). After that I tried to do induction steps but don't know how to proceed or even if I am on the right track.

Photo of steps

I would appreciate any guidance anyone is able to provide. Thank you.

Answer 1

according to your question, we have to find $$\sum_{k=1}^{n} k^{2}(k+1)$$.Let $$S=\sum_{k=1}^{n} k^{2}(k+1)$$. Now $k^{2}(k+1)=k^{3}+k^{2}$. Now just find $$\sum_{k=1}^{n} k^{3}+\sum_{k=1}^{n}k^{2}$$. Now $$\sum_{k=1}^{n} k^{3}=(\frac{n(n+1)}{2})^{2}$$ and $$\sum_{k=1}^{n} k^{2}=(\frac{n(n+1)(2n+1)}{6})$$. now the result of your summation will be $S=(\frac{n(n+1)}{2})^{2}+(\frac{n(n+1)(2n+1)}{6})$. Now take $\frac{n(n+1)}{2}$ common. So, $S=\frac {n(n+1)}{2}(\frac{n(n+1)}{2}+\frac{2n+1}{3})$. So, $S=\frac{n(n+1)}{2}\frac{(3n^{2}+7n+2)}{6}$. So, $S=\frac{n(n+1)(n+2)(3n+1)}{12}$. Hence, proved.

Answer 2

My way: Let $f(x)=\frac14x^4-\frac16x^3-\frac14x^2+\frac16x$. We have $f(x+1)-f(x)=x^3+x^2$.Therefore $\sum_{k=1}^{n}[k^2(k+1)] =\sum_{k=1}^{n}[f(k+1)-f(k)]=f(n+1)-f(1)=\frac{1}{12}n(n+1)(n+2)(3n+1)$