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by-using-the-knowledge-of-sequence-and-series-show-that-the-compound-interest-is-given-by-An-P-1-RT-100-n-

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Question Number 41288 by mondodotto@gmail.com last updated on 04/Aug/18
by using the knowledge of sequence and series show that the compound interest is given by An=P(1+((RT)/(100)))^n
$$\boldsymbol{\mathrm{by}}\:\boldsymbol{\mathrm{using}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{knowledge}}\:\boldsymbol{\mathrm{of}}\: \\ $$$$\boldsymbol{\mathrm{sequence}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{series}}\:\boldsymbol{\mathrm{show}}\:\boldsymbol{\mathrm{that}} \\ $$$$\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{compound}}\:\boldsymbol{\mathrm{interest}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{given}}\:\boldsymbol{\mathrm{by}} \\ $$$$\boldsymbol{\mathrm{An}}=\boldsymbol{\mathrm{P}}\left(\mathrm{1}+\frac{\boldsymbol{\mathrm{RT}}}{\mathrm{100}}\right)^{\boldsymbol{\mathrm{n}}} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 05/Aug/18
begining of the end of year end of year year(principle) − interest − amount 1) P _ ((P×R×1)/(100)) − P+((P×R×1)/(100)) 2) P(1+(R/(100))) − P(1+(R/(100)))×((R×1)/(100)) − P(1+(R/(100))).(R/(100))+P(1+(R/(100))) 3)P(1+(R/(100)))^2 − P(1+(R/(100)))^2 ×(R/(100)) −P{1+(R/(100))}^2 ×(R/(100))+P{1+(R/(100))}^2 thus thus for first year principle(p)=p amount(A)=P(1+(R/(100))) for 2nd year (p)=p(1+(R/(100))) amount(A)=p(1+(R/(100)))^2 for 3rd year(p)=p(1+(R/(100)))^2 amount(A)=p(1+(R/(100)))^3 thus for n year amount(A)=p(1+(R/(100)))^n
$${begining}\:{of}\:{the}\:\:\:\:\:\:\:\:\:\:\:{end}\:{of}\:{year}\:\:\:\:\:{end}\:{of}\:{year} \\ $$$$\:\:{year}\left({principle}\right)\:\:\:\:\:\:\:\:−\:\:\:\:{interest}\:\:\:\:\:\:\:\:\:\:\:\:−\:\:{amount} \\ $$$$\left.\mathrm{1}\right)\:\:\:\:\:\:\:{P}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\_\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{{P}×{R}×\mathrm{1}}{\mathrm{100}}\:\:\:\:\:\:\:\:\:\:−\:\:\:\:\:\:\:\:\:\:{P}+\frac{{P}×{R}×\mathrm{1}}{\mathrm{100}} \\ $$$$\left.\mathrm{2}\right)\:\:\:\:\:\:{P}\left(\mathrm{1}+\frac{{R}}{\mathrm{100}}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{P}\left(\mathrm{1}+\frac{{R}}{\mathrm{100}}\right)×\frac{{R}×\mathrm{1}}{\mathrm{100}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{P}\left(\mathrm{1}+\frac{{R}}{\mathrm{100}}\right).\frac{{R}}{\mathrm{100}}+{P}\left(\mathrm{1}+\frac{{R}}{\mathrm{100}}\right) \\ $$$$\left.\mathrm{3}\right){P}\left(\mathrm{1}+\frac{{R}}{\mathrm{100}}\right)^{\mathrm{2}} \:\:\:\:\:\:\:\:\:\:\:\:−\:\:\:\:\:{P}\left(\mathrm{1}+\frac{{R}}{\mathrm{100}}\right)^{\mathrm{2}} ×\frac{{R}}{\mathrm{100}}\:\:\:\:−{P}\left\{\mathrm{1}+\frac{{R}}{\mathrm{100}}\right\}^{\mathrm{2}} ×\frac{{R}}{\mathrm{100}}+{P}\left\{\mathrm{1}+\frac{{R}}{\mathrm{100}}\right\}^{\mathrm{2}} \\ $$$${thus} \\ $$$${thus}\:{for}\:{first}\:{year}\:{principle}\left({p}\right)={p}\:\:\:\:{amount}\left({A}\right)={P}\left(\mathrm{1}+\frac{{R}}{\mathrm{100}}\right) \\ $$$${for}\:\mathrm{2}{nd}\:{year}\:\left({p}\right)={p}\left(\mathrm{1}+\frac{{R}}{\mathrm{100}}\right)\:\:\:\:\:\:{amount}\left({A}\right)={p}\left(\mathrm{1}+\frac{{R}}{\mathrm{100}}\right)^{\mathrm{2}} \\ $$$${for}\:\mathrm{3}{rd}\:{year}\left({p}\right)={p}\left(\mathrm{1}+\frac{{R}}{\mathrm{100}}\right)^{\mathrm{2}} \:\:\:{amount}\left({A}\right)={p}\left(\mathrm{1}+\frac{{R}}{\mathrm{100}}\right)^{\mathrm{3}} \\ $$$$ \\ $$$${thus}\:{for}\:\:{n}\:\:{year}\:{amount}\left({A}\right)={p}\left(\mathrm{1}+\frac{{R}}{\mathrm{100}}\right)^{{n}} \\ $$

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