Find the summation of the following sequence

Question

Please give me an idea on how to proceed as I am really stuck with this, I have not encountered this type of question before yet my friend gave this summation to me and I am stuck. $${4\choose 1} + \frac{5\choose2}{2}+\frac{6\choose3}{4}+\dots$$

Answer 1

Oh.

I misread the denominator as $n+1$, not $2^n$.

So this is easy.

Note that $\binom{n}{n-3} =\binom{n}{3}$.

Then look at $(1-x)^{-a}$ for $a=4$.

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Here's more.

Your series is

$\sum_{n=1}^{\infty} \frac1{2^{n-1}}\binom{n+3}{n} =2\sum_{n=1}^{\infty} \frac1{2^n}\binom{n+3}{3} $.

Consider $f_a(x) =(1-x)^{-a} =\sum_{n=0}^{\infty} \binom{a+n-1}{n} x^n $.

(see https://en.wikipedia.org/wiki/Binomial_theorem#Newton.27s_generalized_binomial_theorem)

Your series is

$\begin{array}\\ 2\sum_{n=1}^{\infty} \frac1{2^n}\binom{n+3}{n} &=2(-1+\sum_{n=0}^{\infty} \frac1{2^n}\binom{n+3}{n})\\ &=2(f_4(\frac12)-1)\\ &=2(\dfrac1{(1-\frac12)^4}-1)\\ &=2(\dfrac1{\frac12^4}-1)\\ &=30\\ \end{array} $

Answer 2

$$\sum_{r=1}^\infty\binom{r+3}r\left(\dfrac12\right)^{r-1}=\sum_{r=1}^\infty\binom{r+3}3\left(\dfrac12\right)^{r-1}$$

$$=\dfrac26\sum_{r=1}^\infty\left(r^3\left(\dfrac12\right)^r+6r^2\left(\dfrac12\right)^r+11r\left(\dfrac12\right)^r+6\left(\dfrac12\right)^r\right)$$

Now for $|x|<1,$ $$\sum_{r=0}^\infty ax^r=\dfrac a{1-x}$$

We need repeated differentiation

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \sum_{n = 1}^{\infty}{{n + 3 \choose n} \over 2^{n - 1}} & = -2 + 2\sum_{n = 0}^{\infty}{1 \over 2^{n}}\ \overbrace{{-\bracks{n + 3} + n - 1 \choose n}\pars{-1}^{n}}^{\ds{\mbox{Binomial Negation}}}\ =\ -2 + 2\sum_{n = 0}^{\infty} {-4 \choose n}\pars{-\,{1 \over 2}}^{n} \\[5mm] & = -2 + 2\bracks{1 + \pars{-\,{1 \over 2}}}^{-4} = \bbx{\ds{30}} \end{align}

Answer 4

Let $\displaystyle\binom{r+3}r\left(\dfrac12\right)^{r-1}=\binom{r+3}3\left(\dfrac12\right)^{r-1}=f(r+1)-f(r)$

where $f(m)=\left(\dfrac12\right)^m\sum_{r=0}^na_rm^r$ where $a_r$ are arbitrary constants

$\dfrac16(r+3)(r+2)(r+1)\left(\dfrac12\right)^{r-1}$ $=\left(\dfrac12\right)^{r+1}\left(a_0+a_1(r+1)+a_2(r+1)^2+a_3(r+1)^3+\cdots\right)-\left(\dfrac12\right)^r\left(a_0+a_1(r)+a_2(r)^2+a_3(r)^3+\cdots\right)$

$\dfrac{r^3+6r^2+11r+6}6$ $=\left(\dfrac12\right)^2\left(a_0+a_1(r+1)+a_2(r+1)^2+a_3(r+1)^3+\cdots\right)-\left(\dfrac12\right)\left(a_0+a_1(r)+a_2(r)^2+a_3(r)^3+\cdots\right)$

Clearly, $a_r=0$ for $r\ge4$

Comparing the coefficients of $r^3,$ $$\dfrac16=\dfrac{a_3}4-\dfrac{a_3}2\iff a_3=?$$

Comparing the coefficients of $r^2,$ $$1=\dfrac{a_2+3a_3}4-\dfrac{a_2}2\iff a_2=?$$

Similarly, comparing the coefficients of $r$ and the constants, we can find $a_1,a_0$

Using Telescoping Series, $$\sum_{r=1}^\infty\binom{r+3}3\left(\dfrac12\right)^{r-1}=-f(1)$$