The general term T_{r+1} in the binomial expansion is given by T_{r+1} = ^{n}C_{r} a^{n-r} b^{r}

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)} = ^{n}C_{r} 1^{n-r} x^{r}

T_{(r+1)} = ^{n}C_{r} x^{r}

The coefficient of (r+1)^{th} term is ^{n}C_{r}

The r^{th} term is given by (r-1)^{th} term

T_{(r+1-1)} = ^{n}C_{r-1} x^{r-1}

T_{r} = ^{n}C_{r-1} x^{r-1}

the coefficient of r^{th} term is ^{n}C_{r-1}

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

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

T_{r-1} = ^{n}C_{r-2} x^{r-2}

the coefficient of (r-1)^{th} term is ^{n}C_{r-2}

Given that the coefficient of (r-1)^{th}, r^{th} and r+1^{th} 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

Adding equation 2 and 3

2n -8r +10 =0

-3n – 8r – 3 =0

= -n = -7

n =7 and r = 3

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