$(1+x)^{20}=\sum_{r=1}^{20}a_rx^r$ where $a_r=\binom{20}{r}$, find the value $\mathop{\sum\sum}_{0\le i<j\le 20}(a_i-a_j)^2$

Question

$(1+x)^{20}=\sum_{r=1}^{20}a_rx^r$ where $a_r=\binom{20}{r}$, find the value $$\mathop{\sum\sum}_{0\le i<j\le 20}(a_i-a_j)^2$$

I first calculated the value of $\mathop{\sum\sum}_{0\le i<j\le 20}(a_j)^2$

$$A=\mathop{\sum\sum}_{0\le i<j\le 20}(a_j)^2=\frac{1}{2}\left(\mathop{\sum\sum}_{0\le i,j\le20}(a_j)^2-\sum_{0\le i\le20}a_i^2\right)=\frac{1}{2}\left(41\binom{40}{20}-\dbinom{40}{20}\right)=20\dbinom{40}{20}$$

Then I calculated the value of $\mathop{\sum\sum}_{0\le i<j\le 20}(a_ia_j)$

$$B=\mathop{\sum\sum}_{0\le i<j\le 20}(a_ia_j)=\frac{1}{2}\left(\mathop{\sum\sum}_{0\le i,j\le20}(a_ia_j)-\sum_{0\le i\le20}a_i^2\right)=\frac{1}{2}\left(2^{40}-\binom{40}{20}\right)$$

The final answer I got was $2A-2B=41\binom{41}{20}-2^{40}$

But given answer is $41\binom{42}{20}-2^{40}$

Answer 1

From $(a_i - a_j)^2 = a_i^2 - 2 a_i a_j + a_j^2$, I'm pretty sure you want $2A - 2B$. But your $A$ and $B$ give $2A - 2B = 41 \binom{40}{20} - 2^{40}$, which is also not the given answer, so there must be error(s) in your $A$ and/or $B$. When I compute $A$, I get $10 \binom{40}{20}$ and when I compute $B$, I get the same value you have. This suggests a factor of $2$ error in your computation of $A$. I compute: $$\begin{align} A &= \sum_{i=0}^{20} \sum_{j=i+1}^{20} a_j^2 \\ &= \frac{1}{2}\left( \sum_{i=0}^{20} \sum_{j=0}^{20} a_j^2 - \sum_{i=0}^{20} a_i^2 \right) \\ &= \frac{1}{2}\left( 21 \binom{40}{20} - \binom{40}{20}\right) & &\text{(contra. your $41 \dots$) }\\ &= 10 \binom{40}{20}, \end{align}$$ getting the expected factor of $2$ discrepancy.

With this new $A$ and the old $B$, we have $$\begin{align} \sum_{i=0}^{20} \sum_{j=i+1}^{20} (a_i - a_j)^2 &= 2A - 2B \\ &= 20 \binom{40}{20} - \left( 2^{40} - \binom{40}{20} \right) \\ &= 21 \binom{40}{20} - 2^{40}. \end{align}$$

The given answer is more than $10$ times larger than this, so I am dubious about its correctness.

Aside: Checking numerically: $$ \begin{align} \sum_{i=0}^{20} \sum_{j=i+1}^{20} (a_i - a_j)^2 &= 1\,795\,265\,477\,444. \\ 21 \binom{40}{20} - 2^{40} &= 1\,795\,265\,477\,444. \\ 41 \binom{41}{20} - 2^{40} &= 9\,934\,774\,798\,244. \\ 41 \binom{42}{20} - 2^{40} &= 19\,965\,944\,276\,444. \end{align}$$

Answer 2

Using Lagrange Identity (that is $$\sum_{i=1}^nx_i^2\sum_{j=1}^ny_i^2=\sum_{i<j}(x_iy_j-x_jy_i)^2+\left(\sum_{i=1}^nx_iy_i\right)^2.)$$, taking $n=20$, $x_r=a_r$ and $y_r=1$ we get $$20\sum_{i=1}^{20}a_i^2=\sum_{1\le i<j\le 20}(a_i-a_j)^2+\left(\sum_{i=1}^{20}a_i\right)^2...(*)$$.

Thus, remembering that $a_r=\binom{20}{r}$, we have $\sum_{i=1}^{20}a_i=2^{20}-1$ and $\sum_{i=1}^{20}a_i^2=\binom{40}{20}-1$.

Can you conclude? (Note that sum $\sum(a_i-a_j)^2$ in $(*)$ begins in 1, while you need that sum begining in 0).