The coefficients of (r−1)^th, r^th and (r+1)^th terms in the expansion of (x+1)ⁿ are in the ratio 1 : 3 : 5. Find both n and r.

According to the question, Coefficient of T_r−1 : Coefficient of T_r : Coefficient of T_r+1 = 1 : 3 : 5

ⁿC_r−2 : ⁿC_r−1 : ⁿC_r = 1 : 3 : 5

⇒ ⁿC_r−2/1 = ⁿC_r−1/3 = ⁿC_r/5     ...(i)

If ⁿC_r−2/1 = ⁿC_r−1/3

⇒ 3 × ⁿC_r−2 = ⁿC_r−1

⇒ 3 × n! / [(r−2)! (n−r+2)!] = n! / [(r−1)! (n−r+1)!]

⇒ 3(r−1) = n−r+2     ...(ii)

Similarly, from ⁿC_r−1/3 = ⁿC_r/5

⇒ 5 × ⁿC_r−1 = 3 × ⁿC_r

⇒ 5 × n! / [(r−1)! (n−r+1)!] = 3 × n! / [r! (n−r)!]

⇒ 5r = 3n − 3r + 3     ...(iii)

Multiplying (ii) by 2 and subtracting from (iii), we get n = 7.

Putting n = 7 in (ii), we get 7 − 4r = −5 ⇒ r = 3.

Hence, n = 7 and r = 3.