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Question Number 80746 by jagoll last updated on 06/Feb/20

$$whatisconstanterminexpansion$$$${(1+3x)}^5{(\frac{3}{x}+1)}^2$$

Commented by jagoll last updated on 06/Feb/20

$$c\left(x\right)=\frac{{(1+3x)}^5{(3+x)}^2}{x^2}$$$$letp\left(x\right)={(1+3x)}^5{(3+x)}^2$$$$p^{\prime }\left(x\right)=15{(3x+1)}^4{(3+x)}^2+6(3+x){(1+3x)}^5$$$$constantterminc\left(x\right)=\frac{p^{″}\left(0\right)}{2!}$$$$mymethodiscorrect?$$

Commented by mathmax by abdo last updated on 06/Feb/20

$$wehave{(3x+1)}^5{\left(\frac{x+3}{x}\right)}^2=\frac{1}{x^2}{(3x+1)}^5(x^2+6x+9)$$$$={(3x+1)}^5+\frac{6}{x}{(3x+1)}^5+\frac{9}{x^2}{(3x+1)}^5$$$$thecostanttermin{(3x+1)}^{5}is1$$$$\frac{6}{x}{(3x+1)}^5=\frac{6}{x}\sum _{k=0}^5{C}_5^k{\left(3x\right)}^k=6\sum _{k=0}^5{3}^k{C}_5^k{x}^{k-1}$$$$termconstant\Rightarrow k=1\Rightarrow c_1=6\times 3^1\times C_5^1=18\times 5=90$$$$\frac{9}{x^2}{(3x+1)}^5=\frac{9}{x^2}\sum _{k=0}^5{C}_5^k{\left(3x\right)}^k=9\sum _{k=0}^5{3}^k{C}_5^k{x}^{k-2}$$$$termconstant\Rightarrow k=2\Rightarrow c_2=9\times 3^2\times C_5^2=81\times \frac{5!}{2!3!}$$$$=81\times \frac{5\times 4}{2}=810\Rightarrow theconstanttermin{(1+3x)}^5{(\frac{3}{x}+1)}^2$$$$is1+90+810=901$$

Answered by ajfour last updated on 06/Feb/20

$$E=\frac{1}{x^2}{(1+3x)}^5{(3+x)}^2$$$$constantterminEissameas$$$$coefficientof{x}^{2}in{x}^{2}E$$$$=1+\left[2\times 3\times 5\left(3\right)\right]+\left[9\times 10\times 9\right]$$$$=1+90+810=901.$$

Commented by john santu last updated on 06/Feb/20

$$p\left(x\right)=\frac{(243x^5+...+90x^2+15x+1)(x^2+6x+9)}{x^2}$$$$theconstantinp\left(x\right)is$$$$\frac{90x^2.9+15x.6x+1.x^2}{x^2}=$$$$810+90+1=901$$

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