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Home  >>  CBSE XI  >>  Math  >>  Binomial Theorem
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The coefficients of the $(r-1)^{th},r^{th}$ and $(r+1)^{th}$ terms in the expansion of $(x+1)^n$ are in the ratio of $1 : 3 : 5$. Find both $n$ and $r$.

$\begin{array}{1 1}(A)\;n=7,r=3\\(B)\;n=3,r=7\\(C)\;n=4,r=8\\(D)\;n=5,r=3\end{array} $

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1 Answer

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Toolbox:
  • General term $T_{k+1}=C(n,k)x^{n-k}$
  • $C(n,r)=\large\frac{n!}{r!(n-r)!}$
General term in the expansion of $(x+1)^n$ is
$T_{k+1}=C(n,k)x^{n-k}$
Putting $T_{k+1}=T_{r-1}$
$k=r+2$
Coefficient of $T_{r-1}$ is $C(n,r-2)$-----(1)
Putting $T_{k+1}=T_r$
$k=r-1$
Coefficient of $T_r$ is $C(n,r-1)$-----(2)
$T_{r+1}=T_{k+1}$
$r=k$
Coefficient of $T_{r+1}$ is $C(n,r)$-----(3)
According to the problem
$C(n,r-2) : C(n,r-1) : C(n,r) =1 : 3 : 5$
$\large\frac{C(n,r-2)}{1}=\frac{C(n,r-1)}{3}=\frac{C(n,r)}{5}$
If $\large\frac{C(n,r-2)}{1}=\frac{C(n,r-1)}{3}$
Or $3C(n,r-2)=C(n,r-1)$
Or $3\large\frac{n!}{(r-2)!(n+2-r)!}=\frac{1}{(r-1)(r-2)!(n+1-r)!}$
$\large\frac{3}{(n+2-r)}=\frac{1}{r-1}$
$3r-3=n+2-r$-----(4)
$4r=n+5$
Again if If $\large\frac{C(n,r-1)}{3}=\frac{C(n,r)}{5}$
Or $5C(n,r-1)=3C(n,r)$
Or $5\large\frac{n!}{(r-1)!(n+1-r)!}=$$3\large\frac{n!}{r!(n-r)!}$
$5r=3(n+1-r)$
$8r=3n+3$----(5)
From (4) and (5)
$2n+10=3n+3$
$3n-2n=10-3$
From (5) $8r=21+3$
$r=3$
$\therefore n=7,r=3$
Hence (A) is the correct answer.
answered Jun 23, 2014 by sreemathi.v
 

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