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Q10.    The coefficients of the (r-1)th , rth and (r + 1)th terms in the expansion of (x+1)^{n} are in the ratio 1 : 3 : 5. Find n and r.

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As we know that the general  (r+1)^{th} term  T_{r+1} in the binomial expansion of  (a+b)^n  is given by 

T_{r+1}=^nC_ra^{n-r}b^r

So,

(r+1)^{th} Term in  the expansion of  (x+1)^{n}:

T_{r+1}=^nC_rx^{n-r}1^r=^nC_rx^{n-r}

r^{th} Term in  the expansion of  (x+1)^{n}:

T_{r}=^nC_{r-1}x^{n-r+1}1^{r-1}=^nC_{r-1}x^{n-r+1}

(r-1)^{th} Term in  the expansion of  (x+1)^{n}:

T_{r-1}=^nC_{r-2}x^{n-r+2}1^{r-2}=^nC_{r-2}x^{n-r+2}

Now, As given in the question,

T_{r-1}:T_r:T_{r+1}=1:3:5

^nC_{r-2}:^nC_{r-1}:^nC_{r}=1:3:5

\frac{n!}{(r-2)!(n-r+2)!}:\frac{n!}{(r-1)!(n-r+1)!}:\frac{n!}{r!(n-r)!}=1:3:5

From here, we get ,

\frac{r-1}{n-r+2}=\frac{1}{3}\:\:and\:\:\frac{r}{n-r+1}=\frac{3}{5}

Which can be written as 

n-4r+5=0\:\:and\:\:3n-8r+3=0

From these equations we get,

n=7\:\:and\:\:r=3  

Posted by

Pankaj Sanodiya

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