How prove this indentity with this sum $\left(\sum_{1\le i

Question

show that: this follow indentity: $$\sum_{1\le i<j\le n}(y_{i+1}+y_{i+2}+\cdots+y_{j})^2=\bigg(\sum_{k=2}^n (k-1)(n+1-k)^2y_k^2\bigg)+ \bigg(\sum_{2 \leq i,j \leq n} 2(i-1)(n+1-j)y_iy_j\bigg) $$

I have try this $$\sum_{1\le i<j\le n}\left(\sum_{k=i+1}^{j}y_{k}\right)^2=\sum_{j=1}^{n}\sum_{i=1}^{j-1}\left(\sum_{k=i+1}^{j}y^2_{k}+2\sum_{i+1\le p<q\le j}y_{p}y_{q}\right)=I_{1}+I_{2}$$ since $$I_{1}=\sum_{j=1}^{n}\sum_{i=1}^{j-1}\sum_{k=i+1}^{j}y^2_{k}=\sum_{k=1}^{n}\left(\sum_{j=k}^{n}\sum_{i=1}^{k-1}y^2_{k}\right)=\sum_{k=1}^{n}(k-1)(n+1-k)y^2_{k}$$ \begin{align*}I_{2}&=2\sum_{j=1}^{n}\sum_{i=1}^{j-1}\sum_{i+1\le p<q\le j}y_{p}y_{q}=2\sum_{j=1}^{n}\sum_{i=1}^{j}\sum_{q=i+1}^{j}\sum_{p=1}^{q}y_{p}y_{q}=2\sum_{q=1}^{n}\sum_{p=1}^{q-1}\sum_{i=1}^{q-1}\sum_{j=q}^{n}y_{p}y_{q}\\ &=2\sum_{q=1}^{n}\sum_{p=1}^{q-1}(n-q+1)qy_{p}y_{q} \end{align*} then I can't prove $I_{1}+I_{2}=RHS$

Can you someone can use without Mathematical induction methods to prove it? Now I take sometimes to prove this indentity,because I don't know this coefficient $(k-1)(n+1-k)^2$ How to find it? can you explain?

this problem is from this: Prove $\left(\sum_{1 \le i <j \le n} |x_i-x_j|\right)^2 \ge (n-1)\sum_{1\le i<j \le n} (x_i-x_j)^2.$

Thank you

Answer 1

Once for each $i\lt k\le j$, a single $y_k^2$ comes from $(y_{i+1}+y_{i+2}+\dots+y_j)^2$. There are $k-1$ such $i$s and $n-k+1$ such $j$s. That gives a total of $(k-1)(n-k+1)$.

Once for each $i\lt k\lt m\le j$, a single $2y_ky_m$ comes from $(y_{i+1}+y_{i+2}+\dots+y_j)^2$. There are $k-1$ such $i$s and $n-m+1$ such $j$s. That gives a total of $(k-1)(n-m+1)$.

Therefore, it seems to me that $$ \begin{align} &\sum_{1\le i\lt j\le n}(y_{i+1}+y_{i+2}+\dots+y_j)^2\\ &=\sum_{k=2}^n(k-1)(n-k+1)\,y_k^2+\sum_{2\le k\lt m\le n}2(k-1)(n-m+1)\,y_ky_m \end{align} $$

Answer 2

This is a comment with a unwieldy formula. Write $$\sum_{1\le i<j\le n}(y_{i+1}+y_{i+2}+\cdots+y_{j})=\sum_{k=1}^na_ky_k$$ (how many $y_k$'s are for each $k$?)