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.

Asked by Abhisek | 1 year ago |  109

##### Solution :-

The general term Tr+1 in the binomial expansion is given by Tr+1 = nCr an-r br

Here the binomial is (1+x)n with a = 1 , b = x and n = n

The (r+1)th term is given by

T(r+1) = nCr 1n-r xr

T(r+1) = nCr xr

The coefficient of (r+1)th term is nCr

The rth term is given by (r-1)th term

T(r+1-1) = nCr-1 xr-1

Tr = nCr-1 xr-1

the coefficient of rth term is nCr-1

For (r-1)th term we will take (r-2)th term

Tr-2+1 = nCr-2 xr-2

Tr-1 = nCr-2 xr-2

the coefficient of (r-1)th term is nCr-2

Given that the coefficient of (r-1)th, rth and r+1th term are in ratio 1:3:5

Therefore,

$$\dfrac{^nC_{r-2}}{^nC_{r-1}}=\dfrac{1}{3}$$ and $$\dfrac{^nC_{r-1}}{^nC_r}=\dfrac{3}{5}$$

$$\dfrac{\dfrac{n!}{(r-2)!(n-r+2)!}}{\dfrac{n!}{(r-1)!(n-r+1)!}}=\dfrac{1}{3}$$

$$\dfrac{n!}{(r-2)!(n-r+2)!}\times \dfrac{(r-1)!(n-r+1)!}{n!}$$

$$=\dfrac{1}{3}$$

By multiplying

$$\dfrac{(r-1)(r-2)!(n-r+1)!}{(r-2)!(n-r+2)!}=\dfrac{1}{3}$$

$$\dfrac{(r-1)(n-r+1)!}{(n-r+2)(n-r+1)!}=\dfrac{1}{3}$$

On simplifying we get,

$$\dfrac{r-1}{n-r+2}=\dfrac{1}{3}$$

3r-3=n-r+2

n-4r+5=0 ...........(1)

$$\dfrac{^nC_{r-1}}{^nC_r}=\dfrac{3}{5}$$

$$\dfrac{n!}{(r-1)!(n-r+1)}\times \dfrac{(r)!(n-r)!}{n!}$$

$$\dfrac{r(r-1)!(n-r)!}{(r-1)!(n-r+1)(n-r)!}$$

$$\dfrac{r}{n-r+1}=\dfrac{3}{5}$$

5r = 3n – 3r + 3

= 8r – 3n – 3 =0………….2

We have 1 and 2 as

n – 4r ± 5 =0…………1

8r – 3n – 3 =0…………….2

Multiplying equation 1 by number 2

2n -8r +10 =0……………….3

2n -8r +10 =0

-3n – 8r – 3 =0

= -n = -7

n =7 and r = 3

Answered by Pragya Singh | 1 year ago

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