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.

The general term T_r+1 in the binomial expansion is given by T_r+1 = ⁿC_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) = ⁿC_r 1^n-r x^r

T_(r+1) = ⁿC_r x^r

The coefficient of (r+1)^th term is ⁿC_r

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

T_(r+1-1) = ⁿC_r-1 x^r-1

T_r = ⁿC_r-1 x^r-1

the coefficient of r^th term is ⁿC_r-1

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

T_r-2+1 = ⁿC_r-2 x^r-2

T_r-1 = ⁿC_r-2 x^r-2

the coefficient of (r-1)^th term is ⁿC_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