Q1-a-solve-for-x-9-x-5-3-x-6-b-write-down-the-first-4-terms-in-the-binomial-expansion-of-1-3x-7-c-the-sum-S-n-of-the-first-n-th-terms-is-given-by-S-n-3-1-2-3-n-find-d-the-c

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Question Number 35304 by Rio Mike last updated on 17/May/18

$$Q1.a)solveforx{9}^x+5\left(3^x\right)=6$$$$b)writedownthefirst4terms$$$$inthebinomialexpansionof{(1-3x)}^7$$$$c)thesum{S}_{n}ofthefirst{n}^{th}terms$$$$isgivenby{S}_n=3(1-{\left(\frac{2}{3}\right)}^n)find$$$$d)thecommonratio$$$$e)thesumtoinfinityoftheprogression$$

Commented by prakash jain last updated on 17/May/18

$$9^x+5\cdot 3^x=6$$$$3^{2x}+5\cdot 3^x-6=0$$$$3^{2x}+6\cdot 3^x-3^x-6=0$$$$3^x(3^x+6)-(3^x+6)=0$$$$(3^x-1)(3^x+6)=0$$$$3^x=-6(notpossiblefofx\in \mathbb{R}$$$$3^x=1\Rightarrow x=0$$$$ansx=0$$

Commented by Rasheed.Sindhi last updated on 18/May/18

$$\mathrm{Welcome}\mathrm{back}\mathrm{SIR}!\mathrm{I}{\mathrm{haven}}^{\prime }\mathrm{t}\mathrm{seen}\mathrm{your}$$$$\mathrm{post}\mathrm{for}\mathrm{long}\mathrm{time}.\mathrm{Are}\mathrm{you}\mathrm{too}\mathrm{busy}$$$$\mathrm{nowadays}?$$

Commented by rahul 19 last updated on 18/May/18

No problem sir ��

Commented by prakash jain last updated on 18/May/18

$$\mathrm{Was}\mathrm{very}\mathrm{busy}\mathrm{in}\mathrm{office}.\mathrm{Could}$$$$\mathrm{only}\mathrm{read}\mathrm{the}\mathrm{post}\mathrm{once}\mathrm{in}\mathrm{a}\mathrm{week}.$$

Answered by MJS last updated on 18/May/18

$${\left(3^x\right)}^2+5\left(3^x\right)-6=0$$$$\left(3^x\right)=-\frac{5}{2}\pm \sqrt{\frac{25}{4}+6}=-\frac{5}{2}\pm \frac{7}{2}$$$$\left(3^x\right)=-6\vee \left(3^x\right)=1$$$$x_2=0$$$$x_1=\frac{\ln 6}{\ln 3}+\frac{(2n+1)\pi }{\ln 3}\mathrm{i};n\in {\mathbb{N}}_0$$$$$$$${(a-b)}^7=\left(\begin{array}{c}7\\ 0\end{array}\right)\times a^7-\left(\begin{array}{c}7\\ 1\end{array}\right)\times a^{6}b+\left(\begin{array}{c}7\\ 2\end{array}\right)\times a^5{b}^2-\left(\begin{array}{c}7\\ 3\end{array}\right)\times a^4{b}^3+\left(\begin{array}{c}7\\ 4\end{array}\right)\times a^3{b}^4-\left(\begin{array}{c}7\\ 5\end{array}\right)\times a^2{b}^5+\left(\begin{array}{c}7\\ 6\end{array}\right)\times {ab}^6-\left(\begin{array}{c}7\\ 7\end{array}\right)\times {ib}^7$$$${(1-3x)}^7=1-21x+189x^2-945x^3+2835x^4-5103x^5+5103x^6-2187x^7$$$$$$$$\mathrm{the}\mathrm{ratio}\mathrm{is}\frac{2}{3}$$$$p_n={\left(\frac{2}{3}\right)}^n$$$$p_n=q^n\Rightarrow S_n=\frac{q^{n+1}-1}{q-1}$$$$S_{\mathrm{\infty }}=\frac{1}{1-q}=3$$

Answered by Rasheed.Sindhi last updated on 27/May/18

$$\left(\mathrm{c}\right)S_{\mathrm{n}}=3(1-{\left(\frac{2}{3}\right)}^n)$$$$\mathrm{First}\mathrm{term}=a=3(1-{\left(\frac{2}{3}\right)}^1)=1$$$$\divideontimes \mathrm{Common}\mathrm{ratio}r$$$$1+r=\frac{S_2}{S_1}=\frac{3(1-{\left(\frac{2}{3}\right)}^2)}{3(1-{\left(\frac{2}{3}\right)}^1)}[\frac{{\mathrm{S}}_2}{{\mathrm{S}}_1}=\frac{a+ar}{a}=1+r]$$$$=\frac{(1-\frac{2}{3})(1+\frac{2}{3})}{(1-\frac{2}{3})}=\frac{5}{3}$$$$\mathrm{r}=\frac{5}{3}-1=\frac{2}{3}$$$$\divideontimes \mathrm{Sum}\mathrm{of}\mathrm{infinite}\mathrm{terms}=\frac{a}{1-r}=\frac{1}{1-\frac{2}{3}}$$$$=\frac{1}{1/3}=3$$

Commented by Rasheed.Sindhi last updated on 27/May/18

$$\mathrm{Corrected}.$$

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