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Basically

$\sum_{i=0}^{33}\binom{99}{3i} $

I have read about this thread.

I am looking for a conventional approach. Like the one we use in $\sum_{i=0}^{n}\binom{2n}{2i} = 2^{2n-1}$

Community
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Devey Singh Rathore
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3 Answers3

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Here is a completely elementary approach, using nothing but fairly basic facts about binomial coefficients and the sum of a finite geometric series. Let

$$S(n)=\sum_{k=0}^n\binom{3n}{3k}\;.$$

Using the fact that

$$\binom{n}k=\binom{n-3}{k-3}+3\binom{n-3}{k-2}+3\binom{n-3}{k-1}+\binom{n-3}k\;,$$

which is easily proved by repeated applications of Pascal’s identity, we can deduce the recurrence

$$\begin{align*} S(n+1)&=\sum_{k=0}^{n+1}\binom{3n+3}{3k}\\ &=\sum_{k=0}^{n+1}\left(\binom{3n}{3k-3}+3\binom{3n}{3k-2}+3\binom{3n}{3k-1}+\binom{3n}{3k}\right)\\ &=2\sum_{k=0}^n\binom{3n}{3k}+3\left(\sum_{k=1}^n\binom{3n}{3k-2}+\sum_{k=1}^n\binom{3n}{3k-1}\right)\\ &=2S(n)+3\left(\sum_{k=0}^{3n}\binom{3n}k-\sum_{k=0}^n\binom{3n}{3k}\right)\\ &=2S(n)+3\left(2^{3n}-S(n)\right)\\ &=3\cdot2^{3n}-S(n)\;. \end{align*}$$

It’s a first-order recurrence, so we can simply ‘unwind’ it:

$$\begin{align*} S(n)&=3\cdot2^{3(n-1)}-S(n-1)\\ &=3\cdot2^{3(n-1)}-\left(3\cdot2^{3(n-2)}-S(n-2)\right)\\ &=3\cdot2^{3(n-1)}-3\cdot2^{3(n-2)}+S(n-2)\\ &=3\cdot2^{3(n-1)}-3\cdot2^{3(n-2)}+\left(3\cdot2^{3(n-3)}-S(n-3)\right)\\ &=3\cdot2^{3(n-1)}-3\cdot2^{3(n-2)}+3\cdot2^{3(n-3)}-S(n-3)\\ &\;\;\vdots\\ &\overset{*}=3\sum_{k=1}^\ell(-1)^{k+1}2^{3(n-k)}+(-1)^\ell S(n-\ell)\\ &\;\;\vdots\\ &=3\sum_{k=1}^n(-1)^{k+1}2^{3(n-k)}+(-1)^nS(0)\\ &=3\sum_{k=0}^{n-1}(-1)^{n+1-k}2^{3k}+(-1)^n\\ &=3(-1)^{n+1}\sum_{k=0}^{n-1}(-8)^k-(-1)^{n+1}\\ &=(-1)^{n+1}\left(\frac{3\left(1-(-8)^n\right)}9-1\right)\\ &=(-1)^{n+1}\left(\frac{-6-3(-8)^n}9\right)\\ &=\frac13(-1)^n\left(2+(-8)^n\right)\\ &=\frac{8^n+2(-1)^n}3\;, \end{align*}$$

Properly speaking the starred step is a matter of pattern recognition rather than actual proof, so we ought to verify the result by induction on $n$, but in this case it’s pretty clear. There are of course other ways to solve the recurrence.

Brian M. Scott
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    Just a nomenclature note: your second identity,$$ \sum_{j=0}^3\binom{3}{j}\binom{n-3}{k-j} =\binom{n}{k} $$ is an instance of Vandermonde's Identity. – robjohn Apr 21 '16 at 13:13
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    (+1) A possible simplification might be to notice that $S(n+1)=3\cdot2^{3n}-S(n)$ is the same as $(T+1)S(n)=3\cdot2^{3n}$ where $TS(n)=S(n+1)$ is the shift operator. Furthermore, $$ \begin{align} (T-8)(T+1)S(n) &=(T-8)2^{3n}\ &=0 \end{align}$$ and this second order recurrence has the solution $$ S(n) =a8^n+b(-1)^n $$ Using the values for $n\in{0,1,2}$, we arrive at the solution. – robjohn Apr 21 '16 at 13:33
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$$ \begin{align} \sum_{3\mid k}\binom{n}{k} &=\sum_{k=0}^n\binom{n}{k}\frac{1+e^{2\pi ik/3}+e^{-2\pi ik/3}}3\\ &=\frac13\left(2^n+\left(1+e^{2\pi i/3}\right)^n+\left(1+e^{-2\pi i/3}\right)^n\right)\\ &=\frac13\left(2^n+e^{\pi in/3}+e^{-\pi in/3}\right)\\ &=\frac13\left(2^n+2\cos\left(\frac{\pi n}3\right)\right)\\ \end{align} $$ Substitute $n\mapsto 3n$ and get $$ \sum_{3\mid k}\binom{3n}{k}=\frac{8^n+2(-1)^n}3 $$

The approach above mirrors the conventional approach for $$ \begin{align} \sum_{2\mid k}\binom{n}{k} &=\sum_{k=0}^n\binom{n}{k}\frac{1+(-1)^k}2\\ &=\frac12\left(2^n+(1-1)^n\right)\\[6pt] &=\frac{2^n+[n=0]}2 \end{align} $$ Substitute $n\mapsto 2n$ and get $$ \sum_{2\mid k}\binom{2n}{k}=\frac{4^n+[n=0]}2 $$

robjohn
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  • One can simply note that $1+e^{2\pi i/3}+e^{4\pi i/3}=0$, so $$\begin{align}1+e^{2\pi i/3}&=-e^{4\pi i/3}\ &=e^{\pi i/3}\end{align}$$ – robjohn Oct 03 '20 at 07:18
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For the last as $u^2=1\implies u=\pm1$

put $x=-1,1$ one by one $$(1+x)^{2n}=\sum_{r=0}^{2n}x^r$$ and add

For the first case as $v^3=1, v=1,\omega,\omega^2$

Put these values in $$(1+x)^{99}=?$$ and add and use $1+\omega+\omega^2=0$

lab bhattacharjee
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