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Find $\sum_{i=0}^{n}\sum_{j=i}^n(\binom{n}{i}+\binom{n}{j})$

My working:

$$\sum_{i=0}^{n}\sum_{j=i}^n \left(\binom{n}{i}+\binom{n}{j}\right) =\frac{\sum_{i=0}^{n}\sum_{j=0}^n \left(\binom{n}{i}+\binom{n}{j}\right)-\sum_{i=0}^{n}\sum_{j=i} \left(\binom{n}{i}+\binom{n}{j}\right)}{2}$$

Now,
$$\sum_{i=0}^{n}\sum_{j=0}^n \left(\binom{n}{i}+\binom{n}{j} \right)=\sum_{i=0}^{n}n\binom{n}{i}+2^n=n2^n+n2^n=n2^{n+1}$$

and,
$$\sum_{i=0}^{n}\sum_{j=i}\left(\binom{n}{i}+\binom{n}{j}\right)=\sum_{i=0}^{n}2\binom{n}{i}=2\cdot 2^n=2^{n+1}$$

This gives us
$$\sum_{i=0}^{n}\sum_{j=i}^n \left(\binom{n}{i}+\binom{n}{j}\right)=\frac{n2^{n+1}-2^{n+1}}{2}=(n-1)2^n$$

But the correct answer is supposed to be $n2^n$ and I can't figure out what is wrong with my solution. It would be great if I could get a hint to find my error.

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    @user49640 Which sum? – 2017-02-22
  • 2
    In the first sum, note that both $i$ and $j$ runs from $0$ to $n$, i.e. $n+1$ elements. So, $$\sum_{i=0}^{n}\sum_{j=0}^n\Bigg(\binom{n}{i}+\binom{n}{j}\Bigg)=(n+1)2^n+(n+1)2^n=2(n+1)\cdot2^n=(n+1)2^{n+1}$$. It's not $n2^{n+1}$. – 2017-02-22
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    @Dragon Thank you so much...you can put this as the answer – 2017-02-22

3 Answers 3

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I think $n2^n$ is not the correct answer. Just check the simple case for $n=1$, which comes out to be \begin{align} &\bigg(\binom{1}{0}+\binom{1}{0}\bigg)+\bigg(\binom{1}{0}+\binom{1}{1}\bigg)+\bigg(\binom{1}{1}+\binom{1}{1}\bigg)\\ &=3 \times \bigg(\binom{1}{0}+\binom{1}{1}\bigg)=3 \times 2=(1+2) \times2^1 \end{align} Here's a trick which may prove fruitful here. $$\sum_{i=0}^n \sum_{j=i}^n a_{ij}=\sum_{j=0}^n \sum_{i=0}^j a_{ij}$$ We use this trick to compute the sum as follows. \begin{align} \sum_{i=0}^{n}\sum_{j=i}^n\bigg(\binom{n}{i}+\binom{n}{j}\bigg)&=\sum_{i=0}^{n}\sum_{j=i}^n\binom{n}{i}+\sum_{i=0}^{n}\sum_{j=i}^n\binom{n}{j}\\ &=\sum_{i=0}^{n}\sum_{j=i}^n\binom{n}{i}+\sum_{j=0}^{n}\sum_{i=0}^j\binom{n}{j}\\ &=\sum_{i=0}^{n}(n-i+1)\binom{n}{i}+\sum_{j=0}^{n}(j+1)\binom{n}{j}\\ &=\sum_{i=0}^{n}(n-i+1)\binom{n}{i}+\sum_{i=0}^{n}(i+1)\binom{n}{i}\\ &=\sum_{i=0}^{n}(n+2)\binom{n}{i}\\ &=(n+2)2^n \end{align}

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Fix an integer $p$ with $0 \leq p \leq n$. Then $\binom{n}{p}$ appears $n-p + 1$ times as a term $\binom{n}{i}$ (when $i = p$ and $j = p, \dots, n$) and $p +1$ times as $\binom{n}{j}$ (for $i = 0, \dots, p$ and $j=p$). Thus it appears $n + 2$ times in all. Therefore $$\sum_{i=0}^n \sum_{j=i}^n \left[ \binom{n}{i} + \binom{n}{j} \right] = (n+2)\sum_{p=0}^n \binom{n}{p} = (n+2)2^n$$

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    Short and nice. +1 – 2017-02-22
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You have made a small calculation mistake.

We have $$\sum_{i=0 }^{n} \sum_{j=0}^{n} \left(\binom{n}{i} + \binom {n}{j}\right) = 2 (n+1)2^n = (n+1)2^{n+1} $$

Putting this value gives us the correct result of $n\times 2^n $. Hope it helps.