Q.N− Find the coefficint of x^4 in the expansion of (1+3x+10x^2 )(x+(1/x))^(10)


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Question Number 27946 by das47955@mail.com last updated on 17/Jan/18

$Q . N -$ $Find the coefficint of x^{4}$ $in the expansion of$ $(1 + 3 x + 10 x^{2}) {(x + \frac{1}{x})}^{10}$
	Answered by Rasheed.Sindhi last updated on 17/Jan/18
	$(1+3x+10x^2 )(x+(1/x))^(10) =(x+(1/x))^(10) +3x(x+(1/x))^(10) +10x^2 (x+(1/x))^(10) The term containing x^4 =The term containing x^4 in (x+(1/x))^(10) +The term containing x^4 in 3x(x+(1/x))^(10) +The term containing x^4 in 10x^2 (x+(1/x))^(10) T_(r+1) = ((n),(r) ) a^(n−r) b^r The term containing x^4 in (x+(1/x))^(10) T_(r+1) = (((10)),(( r)) ) (x)^(10−r) ((1/x))^r = (((10)),(( r)) ) (x)^(10−2r) 10−2r=4⇒r=3 T_(3+1) =T_4 = (((10)),(( 3)) ) x^4 =120x^4 ...(i) The term containing x^4 in 3x(x+(1/x))^(10) T_(r+1) =3x (((10)),(( r)) ) (x)^(10−r) ((1/x))^r T_(r+1) =3 (((10)),(( r)) ) (x)^(10−2r+1) 11−2r=4 r=7/2 As r may not be fraction hence there′s no term of x^4 in 3x(x+(1/x))^(10) ....(ii) The term containing x^4 in 10x^2 (x+(1/x))^(10) T_(r+1) =10x^2 (((10)),(( r)) ) (x)^(10−r) ((1/x))^r =10 (((10)),(( r)) ) (x)^(10−2r+2) 12−2r=4⇒r=4 T_(4+1) =T_5 =10 (((10)),(( 4)) ) x^4 =2100x^4 ..(iii) From (i) ,(ii) & (iii) 120x^4 +2100x^4 =2220x^4$
	$(1 + 3 x + 10 x^{2}) {(x + \frac{1}{x})}^{10}$ $= {(x + \frac{1}{x})}^{10} + 3 x {(x + \frac{1}{x})}^{10} + {10 x}^{2} {(x + \frac{1}{x})}^{10}$ $The term containing x^{4}$ $= The term containing x^{4} in {(x + \frac{1}{x})}^{10}$ $+ The term containing x^{4} in 3 x {(x + \frac{1}{x})}^{10}$ $+ The term containing x^{4} in {10 x}^{2} {(x + \frac{1}{x})}^{10}$ $T_{r + 1} = (\begin{matrix} n \\ r \end{matrix}) a^{n - r} b^{r}$ $The term containing x^{4} in {(x + \frac{1}{x})}^{10}$ $T_{r + 1} = (\begin{matrix} 10 \\ r \end{matrix}) {(x)}^{10 - r} {(\frac{1}{x})}^{r}$ $= (\begin{matrix} 10 \\ r \end{matrix}) {(x)}^{10 - 2 r}$ $10 - 2 r = 4 \Rightarrow r = 3$ $T_{3 + 1} = T_{4} = (\begin{matrix} 10 \\ 3 \end{matrix}) x^{4} = 120 x^{4} . . . (i)$ $The term containing x^{4} in 3 x {(x + \frac{1}{x})}^{10}$ $T_{r + 1} = 3 x (\begin{matrix} 10 \\ r \end{matrix}) {(x)}^{10 - r} {(\frac{1}{x})}^{r}$ $T_{r + 1} = 3 (\begin{matrix} 10 \\ r \end{matrix}) {(x)}^{10 - 2 r + 1}$ $11 - 2 r = 4$ $r = 7 / 2$ $As r may not be fraction hence$ ${there}^{'} s no term of x^{4} in 3 x {(x + \frac{1}{x})}^{10} . . . . (ii)$ $The term containing x^{4} in {10 x}^{2} {(x + \frac{1}{x})}^{10}$ $T_{r + 1} = {10 x}^{2} (\begin{matrix} 10 \\ r \end{matrix}) {(x)}^{10 - r} {(\frac{1}{x})}^{r}$ $= 10 (\begin{matrix} 10 \\ r \end{matrix}) {(x)}^{10 - 2 r + 2}$ $12 - 2 r = 4 \Rightarrow r = 4$ $T_{4 + 1} = T_{5} = 10 (\begin{matrix} 10 \\ 4 \end{matrix}) x^{4} = {2100 x}^{4} . . (iii)$ $From (i), (ii) & (iii)$ $120 x^{4} + {2100 x}^{4} = {2220 x}^{4}$
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