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Question Number 48705 by rahul 19 last updated on 27/Nov/18

Q.1→ Coefficient of a^8 b^4 c^9 d^9 in expansion of (abc+abd+acd+bcd)^(10) =? Q.2→ Coefficient of (1/x) in expansion of (1+x)^n (1+(1/x))^n =? Q.3→ If x^m occurs in expansion of (x+(1/x^2 ))^(2n) , then its coefficient=?

$${Q}.\mathrm{1}\rightarrow \\ $$$${Coefficient}\:{of}\:{a}^{\mathrm{8}} {b}^{\mathrm{4}} {c}^{\mathrm{9}} {d}^{\mathrm{9}} \:{in}\:{expansion} \\ $$$${of}\:\left({abc}+{abd}+{acd}+{bcd}\right)^{\mathrm{10}} \:=? \\ $$$$ \\ $$$${Q}.\mathrm{2}\rightarrow \\ $$$${Coefficient}\:{of}\:\frac{\mathrm{1}}{{x}}\:{in}\:{expansion}\:{of} \\ $$$$\left(\mathrm{1}+{x}\right)^{{n}} \left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{{n}} =? \\ $$$$ \\ $$$${Q}.\mathrm{3}\rightarrow \\ $$$${If}\:{x}^{{m}} \:{occurs}\:{in}\:{expansion}\:{of}\: \\ $$$$\left({x}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)^{\mathrm{2}{n}} ,\:{then}\:{its}\:{coefficient}=? \\ $$

Commented by rahul 19 last updated on 27/Nov/18

On request of Meritguide sir, these are some problems .... Hope u find them interesting!:)

$${On}\:{request}\:{of}\:{Meritguide}\:{sir}, \\ $$$${these}\:{are}\:{some}\:{problems}\:.... \\ $$$$\left.{Hope}\:{u}\:{find}\:{them}\:{interesting}!:\right) \\ $$

Commented by maxmathsup by imad last updated on 27/Nov/18

3) we have (x+(1/x^2 ))^(2n) =x^(2n) (1+(1/x))^(2n) =x^(2n) Σ_(k=0) ^(2n) C_(2n) ^k x^(−k) =Σ_(k=0) ^n C_(2n) ^k x^(2n−k) so the coefficient is λ_m =C_(2n) ^k /2n−k =m ⇒k=2n−m ⇒λ_m =C_(2n) ^(2n−m) =C_(2n) ^m =(((2n)!)/(m!(2n−m)!))

$$\left.\mathrm{3}\right)\:{we}\:{have}\:\left({x}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)^{\mathrm{2}{n}} ={x}^{\mathrm{2}{n}} \left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}{n}} \:={x}^{\mathrm{2}{n}} \sum_{{k}=\mathrm{0}} ^{\mathrm{2}{n}} {C}_{\mathrm{2}{n}} ^{{k}} \:\:{x}^{−{k}} \:=\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{\mathrm{2}{n}} ^{{k}} \:{x}^{\mathrm{2}{n}−{k}} \:{so}\:{the} \\ $$$${coefficient}\:{is}\:\lambda_{{m}} ={C}_{\mathrm{2}{n}} ^{{k}} \:\:/\mathrm{2}{n}−{k}\:={m}\:\Rightarrow{k}=\mathrm{2}{n}−{m}\:\Rightarrow\lambda_{{m}} ={C}_{\mathrm{2}{n}} ^{\mathrm{2}{n}−{m}} \:\:={C}_{\mathrm{2}{n}} ^{{m}} \\ $$$$=\frac{\left(\mathrm{2}{n}\right)!}{{m}!\left(\mathrm{2}{n}−{m}\right)!} \\ $$$$ \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 27/Nov/18

2)(1+x)^n ×(1+x)^n ×x^(−n) x^(−n) (1+x)^(2n) let (r+1)th term of (1+x)^(2n) contains x^(n−1) 2n_c_r x^r so x^r =x^(n−1) so required x^(−n) ×2n_c_(n−1) ×x^(n−1) =(((2n)!)/((n−1)!(n+1)!))x^(−1) so coefficient is (((2n)!)/((n−1)!(n+1)!))

$$\left.\mathrm{2}\right)\left(\mathrm{1}+{x}\right)^{{n}} ×\left(\mathrm{1}+{x}\right)^{{n}} ×{x}^{−{n}} \\ $$$${x}^{−{n}} \left(\mathrm{1}+{x}\right)^{\mathrm{2}{n}} \\ $$$${let}\:\left({r}+\mathrm{1}\right){th}\:{term}\:{of}\:\left(\mathrm{1}+{x}\right)^{\mathrm{2}{n}} {contains}\:{x}^{{n}−\mathrm{1}} \\ $$$$\mathrm{2}{n}_{{c}_{{r}} } {x}^{{r}} \\ $$$${so}\:{x}^{{r}} ={x}^{{n}−\mathrm{1}} \\ $$$${so}\:{required}\:{x}^{−{n}} ×\mathrm{2}{n}_{{c}_{{n}−\mathrm{1}} } ×{x}^{{n}−\mathrm{1}} \\ $$$$=\frac{\left(\mathrm{2}{n}\right)!}{\left({n}−\mathrm{1}\right)!\left({n}+\mathrm{1}\right)!}{x}^{−\mathrm{1}} \:\:{so}\:{coefficient}\:{is}\:\frac{\left(\mathrm{2}{n}\right)!}{\left({n}−\mathrm{1}\right)!\left({n}+\mathrm{1}\right)!} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 27/Nov/18

3)let (r+1)th term contains x^m 2n_c_r (x)^(2n−r) (x^(−2) )^r 2n_c_r (x)^(2n−3r) 2n−3r=m r=((2n−m)/3) so term is 2n_c_((2n−m)/3) (x)^m so coefficient is (((2n)!)/((((2n−m)/3))!(((4n+m)/3))!))

$$\left.\mathrm{3}\right){let}\:\left({r}+\mathrm{1}\right){th}\:{term}\:{contains}\:{x}^{{m}} \\ $$$$\mathrm{2}{n}_{{c}_{{r}} } \left({x}\right)^{\mathrm{2}{n}−{r}} \left({x}^{−\mathrm{2}} \right)^{{r}} \\ $$$$\mathrm{2}{n}_{{c}_{{r}} } \left({x}\right)^{\mathrm{2}{n}−\mathrm{3}{r}} \\ $$$$\mathrm{2}{n}−\mathrm{3}{r}={m} \\ $$$${r}=\frac{\mathrm{2}{n}−{m}}{\mathrm{3}} \\ $$$${so}\:{term}\:{is} \\ $$$$\mathrm{2}{n}_{{c}_{\frac{\mathrm{2}{n}−{m}}{\mathrm{3}}} } \left({x}\right)^{{m}} \\ $$$${so}\:{coefficient}\:{is}\:\frac{\left(\mathrm{2}{n}\right)!}{\left(\frac{\mathrm{2}{n}−{m}}{\mathrm{3}}\right)!\left(\frac{\mathrm{4}{n}+{m}}{\mathrm{3}}\right)!} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 28/Nov/18

1){abcd((1/d)+(1/c)+(1/b)+(1/a))}^(10) =(abcd)^(10) ((1/a)+(1/b)+(1/c)+(1/d))^(10) C(abcd)^(10) ×(1/a^2 )×(1/b^6 )×(1/c)×(1/d) is the required term which contains a^8 b^4 c^9 d^9 now value of C is =((10!)/(2!6!1!1!))=((10!)/(2!6!))

$$\left.\mathrm{1}\right)\left\{{abcd}\left(\frac{\mathrm{1}}{{d}}+\frac{\mathrm{1}}{{c}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{a}}\right)\right\}^{\mathrm{10}} \\ $$$$=\left({abcd}\right)^{\mathrm{10}} \left(\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}+\frac{\mathrm{1}}{{d}}\right)^{\mathrm{10}} \\ $$$${C}\left({abcd}\right)^{\mathrm{10}} ×\frac{\mathrm{1}}{{a}^{\mathrm{2}} }×\frac{\mathrm{1}}{{b}^{\mathrm{6}} }×\frac{\mathrm{1}}{{c}}×\frac{\mathrm{1}}{{d}}\:{is}\:{the}\:{required} \\ $$$${term}\:{which}\:{contains}\:{a}^{\mathrm{8}} {b}^{\mathrm{4}} {c}^{\mathrm{9}} {d}^{\mathrm{9}} \\ $$$${now}\:{value}\:{of}\:{C}\:{is}\:=\frac{\mathrm{10}!}{\mathrm{2}!\mathrm{6}!\mathrm{1}!\mathrm{1}!}=\frac{\mathrm{10}!}{\mathrm{2}!\mathrm{6}!}\: \\ $$$$ \\ $$

Commented by rahul 19 last updated on 28/Nov/18

thanks sir Ur all answers are absolutely correct!

Commented by tanmay.chaudhury50@gmail.com last updated on 28/Nov/18

thank you Rahul...how is your preparation for IIT JEE

$${thank}\:{you}\:{Rahul}...{how}\:{is}\:{your}\:{preparation} \\ $$$${for}\:{IIT}\:{JEE} \\ $$

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