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If $\displaystyle \alpha = \frac{5}{2!\cdot 3}+\frac{5\cdot 7}{3!\cdot 3^2}+\frac{5\cdot 7 \cdot 9}{4!\cdot 3^3}+\cdots \cdots \infty.$

Then value of $\alpha^2 +4\alpha$ is

Try: Let $$S = \frac{5}{2!\cdot 3}+\frac{5\cdot 7}{3!\cdot 3^2}+\frac{5\cdot 7 \cdot 9}{4!\cdot 3^3}+\cdots \cdots $$

$$S+1 = 1+\frac{5}{2!\cdot 3}+\frac{5\cdot 7}{3!\cdot 3^2}+\frac{5\cdot 7 \cdot 9}{4!\cdot 3^3}+\cdots \cdots $$

Now camparing with $$(1+x)^n = 1+nx+\frac{n(n-1)x^2}{2!}+\frac{n(n-1)(n-2)x^3}{6\cdot 3!}+\cdots \cdots$$

So $\displaystyle nx=\frac{5}{6}$ and $\displaystyle \frac{n(n-1)x^2}{2}=\frac{35}{27}$

So $$\frac{nx(nx-x)}{2}=\frac{5}{12}\cdot \frac{5-6x}{6}=\frac{35}{27}$$

So $\displaystyle x=-\frac{41}{18}$ and $\displaystyle n=-\frac{15}{41}$

I am getting $\displaystyle S+1=\bigg(1-\frac{41}{18}\bigg)^{-\frac{15}{41}}$

but answer of $\alpha^2+4\alpha = 23$

which is not possible from my answer. could some help me how can i solve it, thanks

Community
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DXT
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2 Answers2

2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ Indeed, $\ds{\alpha}$ is given by \begin{align} \alpha & \equiv \sum_{n = 1}^{\infty}{\prod_{k = 1}^{n}\pars{2k + 3} \over \pars{n + 1}!\,3^{n}} = \sum_{n = 1}^{\infty}{2^{n}\prod_{k = 1}^{n}\pars{k + 3/2} \over \pars{n + 1}!\,3^{n}} \\[5mm] & = \sum_{n = 1}^{\infty}{\pars{5/2}^{\large \overline{n}} \over \pars{n + 1}!}\pars{2 \over 3}^{n} = \sum_{n = 1}^{\infty}{\Gamma\pars{5/2 + n}/\Gamma\pars{5/2} \over \pars{n + 1}!}\pars{2 \over 3}^{n} \\[5mm] & = \sum_{n = 1}^{\infty}{\pars{n + 3/2}! \over \pars{n + 1}!\pars{3/2}!}\pars{2 \over 3}^{n} = \sum_{n = 1}^{\infty}{1 \over n + 1}{n + 3/2 \choose n} \pars{2 \over 3}^{n} \\[5mm] & = \sum_{n = 1}^{\infty}\pars{\int_{0}^{1}t^{n}\,\dd t} \bracks{{-5/2 \choose n}\pars{-1}^{n}} \pars{2 \over 3}^{n} \\[5mm] & = \int_{0}^{1}\sum_{n = 1}^{\infty}{-5/2 \choose n} \pars{-\,{2 \over 3}\,t}^{n}\,\dd t = \int_{0}^{1}\bracks{\pars{1 - {2 \over 3}\,t}^{-5/2} - 1}\,\dd t \\[5mm] & \implies \bbx{\alpha = 3\root{3} - 2} \implies \bbx{\alpha^{2} + 4\alpha = \color{red}{\large 23}} \end{align}

Felix Marin
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2

Using \begin{align} \prod_{k=1}^{n} (2 k +3) &= 2^{n} \, \prod_{k=1}^{n} \left(k + \frac{3}{2}\right) = \frac{2^{n} \, \left(n + \frac{3}{2}\right)!}{\left(\frac{3}{2}\right)!} \\ (a)_{n} &= \frac{\Gamma(n+a)}{\Gamma(a)} \\ \sum_{n=0}^{\infty} \frac{(a)_{n} \, x^n}{n!} &= (1-x)^{-a} \end{align} then the following is determined: \begin{align} \alpha &= \sum_{k=1}^{\infty} \frac{\prod_{s=1}^{k} (2s + 3) }{(k+1)! \, 3^{k}} \\ &= \frac{1}{\Gamma(5/2)} \, \sum_{k=1}^{\infty} \frac{\Gamma\left(k + \frac{5}{2}\right) }{(k+1)!} \, \left(\frac{2}{3}\right)^{k} \\ &= \frac{3}{2} \, \frac{\Gamma(3/2)}{\Gamma(5/2)} \, \sum_{k=2}^{\infty} \frac{(3/2)_{k}}{k!} \, \left(\frac{2}{3}\right)^{k} \\ &= \left(1 - \frac{2}{3}\right)^{-3/2} - 1 - 1 \\ &= 3^{3/2} - 2. \end{align} Now, $$\alpha^{2} + 4 \alpha = \alpha (\alpha + 4) = (3^{3/2} -2)(3^{3/2}+2) = 3^{3} - 2^{2}.$$

Leucippus
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