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Q.The coefficients of (r−1)(r-1)th, rrth and (r+1)(r+1)th terms in the expansion of (x+1)n(x+1)^n are in the ratio 1:3:51 : 3 : 5. Find nn and rr.

Jammu Kashmir JkboseJammu and Kashmir Board of School Education (Class 11) 2022Subjective· 4mImportance★★★★★
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Writing the (r−1)th, rth, (r+1)th term coefficients as (nr−2),(nr−1),(nr)\binom{n}{r-2},\binom{n}{r-1},\binom{n}{r} and using the consecutive-ratio identity twice gives n=7n=7, r=3r=3.

In the expansion of (x+1)n(x+1)^n, the general term is Tk+1=(nk)xkT_{k+1}=\binom{n}{k}x^k. So:

  • (r−1)(r-1)th term =Tr−1=T_{r-1}, coefficient =(nr−2)=\binom{n}{r-2}
  • rrth term =Tr=T_r, coefficient =(nr−1)=\binom{n}{r-1}
  • (r+1)(r+1)th term =Tr+1=T_{r+1}, coefficient =(nr)=\binom{n}{r}

Given (nr−2):(nr−1):(nr)=1:3:5\binom{n}{r-2} : \binom{n}{r-1} : \binom{n}{r} = 1:3:5.

First ratio: (nr−1)(nr−2)=3\dfrac{\binom{n}{r-1}}{\binom{n}{r-2}} = 3. Using (nk)(nk−1)=n−k+1k\dfrac{\binom{n}{k}}{\binom{n}{k-1}}=\dfrac{n-k+1}{k} with k=r−1k=r-1:

n−r+2r−1=3  ⇒  n−r+2=3r−3  ⇒  n=4r−5...(i)\frac{n-r+2}{r-1} = 3 \;\Rightarrow\; n-r+2 = 3r-3 \;\Rightarrow\; n = 4r-5 \quad \text{...(i)}

Second ratio: (nr)(nr−1)=53\dfrac{\binom{n}{r}}{\binom{n}{r-1}} = \dfrac{5}{3}. With k=rk=r: …

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