1-5-2-9-4-13-8-17-16-

Question Number 106337 by Dwaipayan Shikari last updated on 04/Aug/20

$$1+\frac{5}{2}+\frac{9}{4}+\frac{13}{8}+\frac{17}{16}+\dots \dots$$

Commented by Dwaipayan Shikari last updated on 04/Aug/20

$$1+5x+9x^2+13x^3+17x^4+\dots \dots (x=\frac{1}{2})$$$$S=1+5x+9x^2+13x^3+..$$$$-2xS=-2x-10x^2-18x^3-\dots$$$$x^{2}S=x^2+5x^3+\dots .$$$$S(1-2x+x^2)=1+3x$$$$S=\frac{1+3x}{1-2x+x^2}=\frac{\frac{5}{2}}{\frac{1}{4}}=10(x=\frac{1}{2})$$

Commented by Ar Brandon last updated on 04/Aug/20

Amazing methode��

Commented by Dwaipayan Shikari last updated on 04/Aug/20

I have found this method idea on this book ���� https://drive.google.com/file/d/12J4x021yiK1X38OHkXoomTz_VuBlcIEt/view?usp=drivesdk

Commented by Ar Brandon last updated on 04/Aug/20

��Thanks bro

Answered by JDamian last updated on 04/Aug/20

$$\underset{k=0}{\sum ^{\mathrm{\infty }}}\frac{4k+1}{2^k}=4\underset{k=0}{\sum ^{\mathrm{\infty }}}\frac{k}{2^k}+\underset{k=0}{\sum ^{\mathrm{\infty }}}\frac{1}{2^k}=4\underset{k=0}{\sum ^{\mathrm{\infty }}}\frac{k}{2^k}+2$$$$f\left(x\right)=1+x+x^2+x^3+\cdot \cdot \cdot =\frac{1}{1-x}\mathrm{\forall }\mid x\mid <1$$$$f^{\prime }\left(x\right)=1+2x+3x^2+\cdot \cdot \cdot =\frac{1}{{(1-x)}^2}\mathrm{\forall }\mid x\mid <1$$$$g\left(x\right)=x\cdot f^{\prime }\left(x\right)=x+2x^2+3x^3+\cdot \cdot \cdot =$$$$=\frac{x}{{(1-x)}^2}\mathrm{\forall }\mid x\mid <1$$$$g\left(\frac{1}{2}\right)=\frac{1}{2^1}+\frac{2}{2^2}+\frac{3}{2^3}+\cdot \cdot \cdot =\frac{\frac{1}{2}}{{(1-\frac{1}{2})}^2}=2$$$$$$$$\underset{k=0}{\sum ^{\mathrm{\infty }}}\frac{4k+1}{2^k}=4\cdot 2+2=10$$

Commented by JDamian last updated on 04/Aug/20

Sigma symbol looks small. How can I get a bigger sigma symbol?

Commented by Ar Brandon last updated on 04/Aug/20

$$\mathrm{By}\mathrm{selecting}\mathrm{the}\mathrm{given}\mathrm{symbol}\mathrm{then}\mathrm{touching}\mathrm{the}3\mathrm{small}$$$$\mathrm{dots}\mathrm{at}\mathrm{the}\mathrm{bottom}-\mathrm{right}\mathrm{corner}.\mathrm{There}{\mathrm{you}}^{\prime }\mathrm{ll}\mathrm{find}{\mathrm{A}}^{\text{blacktrinagledown}}\mathrm{and}$$$${\mathrm{A}}^{▴}.\mathrm{You}\mathrm{can}\mathrm{use}\mathrm{them}\mathrm{to}\mathrm{increase}\mathrm{or}\mathrm{decrease}\mathrm{the}$$$$\mathrm{selected}\mathrm{text}.$$

Commented by JDamian last updated on 04/Aug/20

I edited several times the last line of my answer, I increased the size of the sigma symbol and saved... but my answer is unchanged -- it seems to work fine in the editor but the changes are not shown in my answer

Commented by Ar Brandon last updated on 04/Aug/20

You're right, Sir. I too faced the same problem while trying to demonstrate what I was talking about.

Commented by Tinku Tara last updated on 05/Aug/20

$$\mathrm{There}\mathrm{is}\mathrm{bigger}\mathrm{sigma}\mathrm{symbol}$$$$\mathrm{press}\mathrm{green}\mathrm{color}★\mathrm{key}\mathrm{or}$$$$\mathrm{press}\mathrm{green}\int \mathrm{sign}\mathrm{and}$$$$\mathrm{last}\coprod \mathrm{key}.$$$$\mathrm{Several}\mathrm{enhancement}\mathrm{were}\mathrm{made}$$$$\mathrm{in}\mathrm{last}2\mathrm{weeks}\mathrm{which}\mathrm{are}\mathrm{available}$$$$\mathrm{in}\mathrm{offline}\mathrm{editor}\mathrm{but}\mathrm{not}\mathrm{in}\mathrm{forum}$$$$\mathrm{as}\mathrm{we}\mathrm{giving}\mathrm{user}\mathrm{time}\mathrm{to}\mathrm{update}$$$$\mathrm{to}\mathrm{latest}\mathrm{version}:$$$$\bullet \mathrm{Strike}\mathrm{thru}$$$$\bullet \mathrm{underline}$$$$\bullet \mathrm{curly}\mathrm{bracket}\mathrm{below}$$$$\bullet \mathrm{font}\mathrm{size}\mathrm{for}\mathrm{every}\mathrm{character}$$$$\bullet \mathrm{table}$$$$\bullet \mathrm{borderless}\mathrm{table}(\mathrm{this}\mathrm{is}\mathrm{primarily}$$$$\mathrm{intented}\mathrm{for}\mathrm{use}\mathrm{in}\mathrm{writing}\mathrm{multiple}$$$$\mathrm{line}\mathrm{in}\mathrm{drawing})$$

Commented by Tinku Tara last updated on 05/Aug/20

Commented by Tinku Tara last updated on 05/Aug/20

Commented by Tinku Tara last updated on 05/Aug/20

Answered by 1549442205PVT last updated on 04/Aug/20

$$\mathrm{The}\mathrm{general}\mathrm{term}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{sequence}\mathrm{is}$$$$\frac{4\mathrm{n}-3}{2^{\mathrm{n}-1}}.\mathrm{We}\mathrm{need}\mathrm{calculate}\underset{\mathrm{n}=1}{\stackrel{\mathrm{\infty }}{\mathrm{\Sigma }}}\frac{4\mathrm{n}-3}{2^{\mathrm{n}-1}}=4\underset{\mathrm{n}=1}{\stackrel{\mathrm{\infty }}{\mathrm{\Sigma }}}\frac{\mathrm{n}}{2^{\mathrm{n}-1}}-3\underset{\mathrm{n}=1}{\stackrel{\mathrm{\infty }}{\mathrm{\Sigma }}}\frac{1}{2^{\mathrm{n}-1}}$$$$\mathrm{Set}{\mathrm{S}}_{\mathrm{k}}=\underset{\mathrm{n}=1}{\sum ^{\mathrm{k}}}\frac{\mathrm{n}}{2^{\mathrm{n}-1}}=\frac{1}{1}+\frac{2}{2}+\frac{3}{2^2}+\frac{4}{2^3}+\dots +\frac{\mathrm{k}}{2^{\mathrm{k}-1}}$$$$\frac{1}{2}{\mathrm{S}}_{\mathrm{k}}=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+\dots .+\frac{\mathrm{k}-1}{2^{\mathrm{k}-1}}+\frac{\mathrm{k}}{2^{\mathrm{k}}}$$$$\mathrm{Substrating}\mathrm{two}\mathrm{above}\mathrm{equalities}\mathrm{we}\mathrm{get}$$$$0.{5\mathrm{S}}_{\mathrm{k}}=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\dots +\frac{1}{2^{\mathrm{k}-1}}-\frac{\mathrm{k}}{2^{\mathrm{k}}}$$$$=\frac{1-{\left(\frac{1}{2}\right)}^{\mathrm{k}+1}}{\frac{1}{2}}-\frac{\mathrm{k}}{2^{\mathrm{k}}}\Rightarrow {\mathrm{S}}_{\mathrm{k}}=4(1-{\left(\frac{1}{2}\right)}^{\mathrm{k}+1})-\frac{\mathrm{k}}{2^{\mathrm{k}-1}}$$$$\Rightarrow \underset{\mathrm{n}=1}{\stackrel{\mathrm{\infty }}{\mathrm{\Sigma }}}\frac{\mathrm{n}}{2^{\mathrm{n}-1}}=\underset{\mathrm{k}\to \mathrm{\infty }}{\lim }{\mathrm{S}}_{\mathrm{k}}=\underset{\mathrm{k}\to \mathrm{\infty }}{\lim }[4(1-{\left(\frac{1}{2}\right)}^{\mathrm{k}+1})-\frac{\mathrm{k}}{2^{\mathrm{k}-1}}]=4$$$$\mathrm{\Sigma }\frac{1}{2^{\mathrm{n}-1}}=\frac{1}{1}+\frac{1}{2}+\frac{1}{2^2}+\dots =\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }(\frac{1}{1}+\frac{1}{2}+\frac{1}{2^2}+\dots +\frac{1}{2^{\mathrm{n}}})=$$$$\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{(1-{\left(\frac{1}{2}\right)}^{\mathrm{n}+1})}{1-\frac{1}{2}}=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\left[2(1-{\left(\frac{1}{2}\right)}^{\mathrm{n}+1})\right]=2$$$$\mathrm{Hence},\underset{\mathrm{n}=1}{\stackrel{\mathrm{\infty }}{\mathrm{\Sigma }}}\frac{4\mathrm{n}-3}{2^{\mathrm{n}-1}}=4\underset{\mathrm{n}=1}{\stackrel{\mathrm{\infty }}{\mathrm{\Sigma }}}\frac{\mathrm{n}}{2^{\mathrm{n}-1}}-3\underset{\mathrm{n}=1}{\stackrel{\mathrm{\infty }}{\mathrm{\Sigma }}}\frac{1}{2^{\mathrm{n}-1}}$$$$=4.4-3.2=10$$

Commented by Dwaipayan Shikari last updated on 04/Aug/20

$$Oritcanbe?$$$$S=1+\frac{5}{2}+\frac{9}{4}+\frac{13}{8}+\dots$$$$\frac{S}{2}=\frac{1}{2}+\frac{5}{4}+\frac{9}{8}+\dots$$$$\dots .\left(subtracting\right)$$$$\frac{S}{2}=1+4(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\dots .)$$$$\frac{S}{2}=1+4\left(\frac{\frac{1}{2}}{1-\frac{1}{2}}\right)=10$$

Answered by mathmax by abdo last updated on 04/Aug/20

$$\mathrm{S}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{4\mathrm{n}+1}{2^{\mathrm{n}}}\Rightarrow \mathrm{S}=4\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{\mathrm{n}}{2^{\mathrm{n}}}+\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{1}{2^{\mathrm{n}}}\mathrm{we}\mathrm{have}$$$$\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{1}{2^{\mathrm{n}}}=\frac{1}{1-\frac{1}{2}}=2$$$$\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{\mathrm{n}}{2^{\mathrm{n}}}=\mathrm{w}\left(\frac{1}{2}\right)\mathrm{with}\mathrm{w}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{nx}}^{\mathrm{n}}(\mathrm{we}\mathrm{take}\mid \mathrm{x}\mid <1)$$$$\mathrm{we}\mathrm{have}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{x}}^{\mathrm{n}}=\frac{1}{1-\mathrm{x}}\Rightarrow \sum _{\mathrm{n}=1}^{\mathrm{\infty }}{\mathrm{nx}}^{\mathrm{n}-1}=\frac{1}{{(1-\mathrm{x})}^2}\Rightarrow$$$$\sum _{\mathrm{n}=1}^{\mathrm{\infty }}{\mathrm{nx}}^{\mathrm{n}}=\frac{\mathrm{x}}{{(1-\mathrm{x})}^2}=\mathrm{w}\left(\mathrm{x}\right)\Rightarrow \mathrm{w}\left(\frac{1}{2}\right)=\frac{1}{2{(1-\frac{1}{2})}^2}=\frac{4}{2}=2\Rightarrow$$$$\mathrm{S}=4.2+2=10$$

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