We have known that: \(\sum u(x)\Delta v(x) = u(x)v(x) - \sum Ev(x)\Delta u(x)\)
So apply the formula to our problem: \(\sum_{1\le k \le n}k^2H_{n+k}\)
Let \(u(x) = H_{n+x}\) and \(\Delta v(x) = x^2\)
We can get that: \(\Delta u(x) = \dfrac{1}{n+x+1}\)， \(v(x) = \dfrac{1}{3}x(x-1)(x-2) + \dfrac{1}{2}x(x-1) = \dfrac{1}{6}x(x-1)(2x-1)\)
so :
\(\sum x^2H_{n+x} = \dfrac{1}{6}x(x-1)(2x-1)H_{n+x} - \sum_{0 \le k \le n}\dfrac{1}{6}x(x+1)(2x+1)*\dfrac{1}{n+x+1}\)
Observe the above formula, to solve this problem, we need get the closed form of \(\sum_{0 \le k \le n}\dfrac{1}{6}x(x+1)(2x+1)*\dfrac{1}{n+x+1}\), this is our next step.
First of all ,w need to simply the general term formula : \(\dfrac{1}{6}x(x+1)(2x+1)*\dfrac{1}{n+x+1}\). Here is the steps how I handle it:
\[\begin{align*}
&\dfrac{1}{6}x(x+1)(2x+1)*\dfrac{1}{n+x+1}\\ &= \dfrac{1}{6}x[(x+1+n)-n](2x+1)*\dfrac{1}{n+x+1} \\
&= \dfrac{1}{6}(x(2x+1) - \dfrac{nx(2x+1)}{n+x+1}) \\
&= \dfrac{1}{6}(x(2x+1) - \dfrac{nx(2x+1+1+2n -(1+2n))}{n+x+1}) \\
&= \dfrac{1}{6}(x(2x+1) - \dfrac{nx(2(n+x+1) -(1+2n))}{n+x+1}) \\
&= \dfrac{1}{6}(x(2x+1) - 2nx + \dfrac{n(1+2n)x}{n+x+1}) \\
&= \dfrac{1}{6}(x(2x+1) - 2nx + \dfrac{n(1+2n)(x+n+1-(n+1))}{n+x+1}) \\
&= \dfrac{1}{6}(x(2x+1) - 2nx + n(2n+1) -\dfrac{n(1+2n)(n+1)}{n+x+1}) \\
&= \dfrac{1}{6}x(2x+1) - \dfrac{1}{3}nx +\dfrac{1}{6}n(2n+1)-\dfrac{1}{6}n(n+1)(2n+1)*\dfrac{1}{n+x+1}
\end{align*}
\]
Now, we can easily get the sum of \(\dfrac{1}{6}x(x+1)(2x+1)*\dfrac{1}{n+x+1}\) from 1 to n:
\[\begin{align*}
&\sum_{1 \le x \le n}\dfrac{1}{6}x(x+1)(2x+1)*\dfrac{1}{n+x+1}\\
&= \sum_{1 \le x \le n}[\dfrac{1}{6}x(2x+1) - \dfrac{1}{3}nx +\dfrac{1}{6}n(2n+1)-\dfrac{1}{6}n(n+1)(2n+1)*\dfrac{1}{n+x+1}
]\\
&= \dfrac{1}{6}\sum_{1 \le x \le n}x(2x+1) - \dfrac{1}{3}n\sum_{1 \le x \le n}x +\dfrac{1}{6}\sum_{1 \le x \le n}n(2n+1)-\dfrac{1}{6}n(n+1)(2n+1)*\sum_{1 \le x \le n}\dfrac{1}{n+x+1} \\
&= \dfrac{1}{3}\sum_{1 \le x \le n}x^2 + \dfrac{1}{6}\sum_{1 \le x \le n}x -\dfrac{1}{3}n\sum_{1 \le x \le n}x + \dfrac{1}{6}n^2(2n+1) - \dfrac{1}{6}n(n+1)(2n+1)*(H_{2n+1} -H_{n+1})\\
&=\dfrac{1}{18}n(n+1)(2n+1) + \dfrac{1}{12}n(n+1) -\dfrac{1}{6}n^2(n+1)+ \dfrac{1}{6}n^2(2n+1)- \dfrac{1}{6}n(n+1)(2n+1)*(H_{2n+1} -H_{n+1})\\
&=\dfrac{1}{18}n(n+1)(2n+1) + \dfrac{1}{12}n(n+1) + \dfrac{1}{6}n^3 - \dfrac{1}{6}n(n+1)(2n+1)*(H_{2n+1} -H_{n+1})\\
&= \dfrac{1}{36}n(n+1)(4n+5) + \dfrac{1}{6}n^3 -\dfrac{1}{6}n(n+1)(2n+1)*(H_{2n+1} -H_{n+1})\\
\end{align*}
\]
3、The Most Existing Moment !!!
As we have got the two parts of the final result, now we just need to sum them, and merge the expression carefully.
\[\begin{align*}
\sum_{1\le k \le n}k^2H_{n+k} &= \sum_{1 \le k < n+1}k^2H_{n+k}\\
&= \sum_1^{n+1}k^2H_{n+k}\delta k\\
&= \dfrac{1}{6}k(k-1)(2k-1)H_{n+k}|_1^{n+1} - \sum_1^{n+1}(\dfrac{1}{6}k(k+1)(2k+1)*\dfrac{1}{n+k+1}) \delta k\\
&=\dfrac{1}{6}k(k-1)(2k-1)H_{n+k}|_1^{n+1} - \sum_{1 \le k \le n}\dfrac{1}{6}k(k+1)(2k+1)*\dfrac{1}{n+k+1}\\
&= \dfrac{1}{6}n(n+1)(2n+1)H_{2n+1} - \dfrac{1}{36}n(n+1)(4n+5) - \dfrac{1}{6}n^3 + \dfrac{1}{6}n(n+1)(2n+1)*(H_{2n+1} -H_{n+1})\\
&= \dfrac{1}{6}n(n+1)(2n+1)(H_{2n+1} -H_{n+1}) - \dfrac{1}{36}n(n+1)(4n+5) - \dfrac{1}{6}n^3 \\
&= \dfrac{1}{6}n(n+1)(2n+1)(2H_{2n} + \dfrac{2}{2n+1} - H_n - \dfrac{1}{n+1})- \dfrac{1}{36}n(n+1)(4n+5) - \dfrac{1}{6}n^3 \\
&= \dfrac{1}{6}n(n+1)(2n+1)(2H_{2n} - H_n) + \dfrac{1}{6}n- \dfrac{1}{36}n(n+1)(4n+5) - \dfrac{1}{6}n^3 \\
&= \dfrac{1}{6}n(n+1)(2n+1)(2H_{2n} - H_n) - \dfrac{1}{36}n(10n^2+9n-1)
\end{align*}
\]
Finally we got the result :
\[\begin{align*}
\sum_{1\le k \le n}k^2H_{n+k}
&= \dfrac{1}{6}n(n+1)(2n+1)(2H_{2n} - H_n) - \dfrac{1}{36}n(10n^2+9n-1)
\end{align*}
\]
