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Question-124247

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Question Number 124247 by ZiYangLee last updated on 02/Dec/20
Commented by ZiYangLee last updated on 02/Dec/20
8 and 9, help sir...
$$\mathrm{8}\:\mathrm{and}\:\mathrm{9},\:\mathrm{help}\:\mathrm{sir}… \\ $$
Answered by bramlexs22 last updated on 02/Dec/20
(8)•(x^2 +(1/x))^n = Σ_(r=0) ^n ((n),(r) ) (x^(2(n−r)) )(x^(−r) ) T_4 = ((n),(3) ) x^(2n−6) .x^(−3) = ((n.(n−1).(n−2))/6) x^(9−n) T_(13) = ((n),((12)) ) x^(2n−24) .x^(−12) = ((n!)/(12!(n−12)!)).x^(36−n) coefficient of 4^(th) and 13^(th) are equal then ((n!)/(3!(n−3)!)) = ((n!)/(12!(n−12)!)) ⇔ 3!(n−3)(n−4)...(n−12)!=12.11...3!(n−12)! we get n=15. recall formula ((n),(r) ) = ((( n)),((n−r)) ) we want to find the term indenpendent of x in the expansion (x^2 +(1/x))^(15) (x^2 +(1/x))^(15) =Σ_(r=0) ^(15) (((15)),(( r)) ) x^(30−2r) .(x^(−r) ) the term independent of x we get if x^(30−3r) = x^0 ; r = 10. Thus the term independent of x = (((15)),((10)) ) = ((15.14.13.12.11)/(5.4.3.2.1))=3003 ((15×14×13×12×11)/(5×4×3×2×1)) 3003.0
$$\left(\mathrm{8}\right)\bullet\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}}\right)^{{n}} =\:\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}\begin{pmatrix}{{n}}\\{{r}}\end{pmatrix}\:\left({x}^{\mathrm{2}\left({n}−{r}\right)} \right)\left({x}^{−{r}} \right) \\ $$$${T}_{\mathrm{4}} \:=\:\begin{pmatrix}{{n}}\\{\mathrm{3}}\end{pmatrix}\:{x}^{\mathrm{2}{n}−\mathrm{6}} .{x}^{−\mathrm{3}} \:=\:\frac{{n}.\left({n}−\mathrm{1}\right).\left({n}−\mathrm{2}\right)}{\mathrm{6}}\:{x}^{\mathrm{9}−{n}} \\ $$$${T}_{\mathrm{13}} =\begin{pmatrix}{{n}}\\{\mathrm{12}}\end{pmatrix}\:{x}^{\mathrm{2}{n}−\mathrm{24}} .{x}^{−\mathrm{12}} \:=\:\frac{{n}!}{\mathrm{12}!\left({n}−\mathrm{12}\right)!}.{x}^{\mathrm{36}−{n}} \\ $$$${coefficient}\:{of}\:\mathrm{4}^{{th}} \:{and}\:\mathrm{13}^{{th}} \:{are}\:{equal} \\ $$$${then}\:\frac{{n}!}{\mathrm{3}!\left({n}−\mathrm{3}\right)!}\:=\:\frac{{n}!}{\mathrm{12}!\left({n}−\mathrm{12}\right)!} \\ $$$$\Leftrightarrow\:\mathrm{3}!\left({n}−\mathrm{3}\right)\left({n}−\mathrm{4}\right)…\left({n}−\mathrm{12}\right)!=\mathrm{12}.\mathrm{11}…\mathrm{3}!\left({n}−\mathrm{12}\right)! \\ $$$${we}\:{get}\:{n}=\mathrm{15}.\:{recall}\:{formula}\: \\ $$$$\begin{pmatrix}{{n}}\\{{r}}\end{pmatrix}\:=\:\begin{pmatrix}{\:\:\:{n}}\\{{n}−{r}}\end{pmatrix} \\ $$$${we}\:{want}\:{to}\:{find}\:{the}\:{term}\:{indenpendent} \\ $$$${of}\:{x}\:{in}\:{the}\:{expansion}\:\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}}\right)^{\mathrm{15}} \\ $$$$\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}}\right)^{\mathrm{15}} =\underset{{r}=\mathrm{0}} {\overset{\mathrm{15}} {\sum}}\begin{pmatrix}{\mathrm{15}}\\{\:\:{r}}\end{pmatrix}\:{x}^{\mathrm{30}−\mathrm{2}{r}} .\left({x}^{−{r}} \right) \\ $$$${the}\:{term}\:{independent}\:{of}\:{x}\:{we}\:{get}\:{if} \\ $$$${x}^{\mathrm{30}−\mathrm{3}{r}} \:=\:{x}^{\mathrm{0}} \:;\:{r}\:=\:\mathrm{10}.\:{Thus}\:{the}\:{term} \\ $$$${independent}\:{of}\:{x}\:=\:\begin{pmatrix}{\mathrm{15}}\\{\mathrm{10}}\end{pmatrix}\:=\:\frac{\mathrm{15}.\mathrm{14}.\mathrm{13}.\mathrm{12}.\mathrm{11}}{\mathrm{5}.\mathrm{4}.\mathrm{3}.\mathrm{2}.\mathrm{1}}=\mathrm{3003} \\ $$$$\frac{\mathrm{15}×\mathrm{14}×\mathrm{13}×\mathrm{12}×\mathrm{11}}{\mathrm{5}×\mathrm{4}×\mathrm{3}×\mathrm{2}×\mathrm{1}} \\ $$$$\mathrm{3003}.\mathrm{0} \\ $$$$ \\ $$
Commented by ZiYangLee last updated on 02/Dec/20
wow thanks!
$$\mathrm{wow}\:\mathrm{thanks}! \\ $$
Answered by bramlexs22 last updated on 02/Dec/20
(9)•• (1+x)^(18) =Σ_(r=0) ^(18) (((18)),(( r)) ) x^(18−r) coefficient of the (2r+4)^(th) = ((( 18)),((2r+3)) ) = ((18!)/((2r+3)!(15−2r)!)) coefficient of (r−2)^(th) = ((( 18)),((r−3)) ) =((18!)/((r−3)!(21−r)!)) the (2r+3)!(15−2r)!=(r−3)!(21−r)! we get r = 6.
$$\left(\mathrm{9}\right)\bullet\bullet\:\left(\mathrm{1}+{x}\right)^{\mathrm{18}} =\underset{{r}=\mathrm{0}} {\overset{\mathrm{18}} {\sum}}\begin{pmatrix}{\mathrm{18}}\\{\:\:{r}}\end{pmatrix}\:{x}^{\mathrm{18}−{r}} \\ $$$${coefficient}\:{of}\:{the}\:\left(\mathrm{2}{r}+\mathrm{4}\right)^{{th}} \:=\:\begin{pmatrix}{\:\:\:\:\mathrm{18}}\\{\mathrm{2}{r}+\mathrm{3}}\end{pmatrix}\:=\:\frac{\mathrm{18}!}{\left(\mathrm{2}{r}+\mathrm{3}\right)!\left(\mathrm{15}−\mathrm{2}{r}\right)!} \\ $$$${coefficient}\:{of}\:\left({r}−\mathrm{2}\right)^{{th}} =\begin{pmatrix}{\:\:\mathrm{18}}\\{{r}−\mathrm{3}}\end{pmatrix}\:=\frac{\mathrm{18}!}{\left({r}−\mathrm{3}\right)!\left(\mathrm{21}−{r}\right)!} \\ $$$${the}\:\left(\mathrm{2}{r}+\mathrm{3}\right)!\left(\mathrm{15}−\mathrm{2}{r}\right)!=\left({r}−\mathrm{3}\right)!\left(\mathrm{21}−{r}\right)! \\ $$$${we}\:{get}\:{r}\:=\:\mathrm{6}.\: \\ $$

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