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Evaluation of sum $\displaystyle \sum{\hspace{-0.3 cm}\sum_{0\leq i<j \leq n}}j\cdot \binom{n}{i}$

$\bf{My\: Try::}$ We can write it as $$\displaystyle 1\cdot \binom{n}{0}+2\cdot \left[\binom{n}{0}+\binom{n}{1}\right]+3\cdot \left[\binom{n}{0}+\binom{n}{1}+\binom{n}{2}\right]+.....+n\cdot \left[\binom{n}{0}+\binom{n}{1}+\binom{n}{2}......+\binom{n}{n}\right]$$

Now I did not understand How can i solve after that, Help me

Thanks.

If there is any combinational prove, Then plz explain here, Thanks

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1 Answer 1

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First reverse the order of summation:

$$\sum_{j=1}^n\sum_{i=0}^{j-1}j\binom{n}i=\sum_{i=0}^{n-1}\binom{n}i\sum_{j=i+1}^nj\;.$$

Now

$$\sum_{j=i+1}^nj=\frac{(n-i)(n+i+1)}2\;,$$

and

$$\binom{n}i(n-i)=\binom{n}{n-i}(n-i)=n\binom{n-1}{n-i-1}\;,$$

so your expression is equal to

$$\begin{align*} \frac{n}2\sum_{i=0}^{n-1}(n+i+1)\binom{n-1}{n-1-i}&=\frac{n}2\left((n+1)\sum_{i=0}^{n-1}\binom{n-1}i+\sum_{i=0}^{n-1}i\binom{n-1}i\right)\\ &=2^{n-1}\binom{n+1}2+\frac{n}2\sum_{i=0}^{n-1}i\binom{n-1}i\;; \end{align*}$$

can you finish off that last summation?

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