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Question Number 94319 by Mozay1 last updated on 18/May/20

Commented by john santu last updated on 18/May/20

$$\left(1\mathrm{b}\right){(2+3\mathrm{y}+{\mathrm{y}}^2)}^5={\left\{(\mathrm{y}+1)(\mathrm{y}+2)\right\}}^5$$$$=\left\{\underset{\mathrm{k}=0}{\sum ^5}\mathrm{C}{}_{\mathrm{k}}^5{\mathrm{y}}^{5-\mathrm{k}}{1}^{\mathrm{k}}\times \underset{\mathrm{i}=0}{\sum ^5}{\mathrm{C}}_{\mathrm{i}}^5{\mathrm{y}}^{5-\mathrm{i}}{2}^{\mathrm{i}}\right\}$$$${\mathrm{y}}^3={\mathrm{y}}^{5-\mathrm{k}}\times {\mathrm{y}}^{5-\mathrm{i}}={\mathrm{y}}^{10-(\mathrm{k}+\mathrm{i})}$$$$\mathrm{k}+\mathrm{i}=7;0⩽\mathrm{k}+\mathrm{i}⩽10$$$$\mathrm{k}=2,\mathrm{i}=5\Rightarrow {\mathrm{C}}_2^5.{\mathrm{C}}_5^5.32=320$$$$\mathrm{k}=3,\mathrm{i}=4\Rightarrow {\mathrm{C}}_3^5.{\mathrm{C}}_4^5.16=800$$$$\mathrm{k}=4,\mathrm{i}=3\Rightarrow {\mathrm{C}}_4^5.{\mathrm{C}}_3^5.8=400$$$$\mathrm{k}=5,\mathrm{i}=2\Rightarrow {\mathrm{C}}_5^5.{\mathrm{C}}_2^5.4=40$$$$\mathrm{so}\mathrm{coeff}\mathrm{of}{\mathrm{y}}^3\mathrm{is}1,560$$

Commented by mathmax by abdo last updated on 18/May/20

$$1){(x+2)}^5=\sum _{k=0}^5{C}_5^k{x}^k.2^{5-k}$$$$=C_5^0{2}^5+C_5^{1}x.2^4+C_5^2{x}^2.2^3+C_5^3{x}^3.2^2+C_5^4{x}^4.2+C_5^5{x}^5$$$$=32+16.5x+8C_5^2{x}^2+4C_5^3{x}^3+2C_5^4{x}^4+x^5$$$$C_5^2=\frac{5!}{2!3!}=\frac{5.4}{2}=10=C_5^3,C_5^4=5\Rightarrow$$$${(x+2)}^5=32+80x+80x^2+40x^3+10x^4+x^5$$

Commented by mathmax by abdo last updated on 18/May/20

$$Q2i){(2x+3)}^5=\sum _{k=0}^5{C}_5^k{\left(2x\right)}^k.3^{5-k}$$$$=C_5^0.3^5+C_5^1\left(2x\right).3^4+C_5^2{\left(2x\right)}^2.3^3+C_5^3{\left(2x\right)}^3.3^2+C_5^4{\left(2x\right)}^4.3+C_5^5{\left(2x\right)}^5$$$$=3^5+2.3^4{C}_5^{1}x+4.3^3{C}_5^2{x}^2+8.3^2{C}_5^3{x}^3+16.3C_5^4{x}^4+32x^5$$$$=3^5+32\times 5x+40.3^3{x}^2+80.3^2{x}^3+48\times 5x^4+32x^5$$$${(3-2x)}^5=3^5-32\times 5x+40.3^3{x}^2-809x^3+48\times 5x^4-32x^5\Rightarrow$$$${(3+2x)}^5+{(3-2x)}^5=2(3^5+40\times 27x^2+48\times 5x^4)$$$$=2\times 3^5+80\times 27x^2+96\times 5x^4$$

Commented by mathmax by abdo last updated on 18/May/20

$$Q_{3}b)letf\left(x\right)=(x+1){(1-2x)}^5\Rightarrow f\left(x\right)\sim f\left(0\right)+f^{{}^{\prime }}\left(0\right)x$$$$f\left(0\right)=1and{f}^{{}^{\prime }}\left(x\right)={(1-2x)}^5-10(x+1){(1-2x)}^4$$$$\Rightarrow f^{{}^{\prime }}\left(0\right)=1-10=-9\Rightarrow f\left(x\right)\sim 1-9x$$

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