The coefficient 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 both n and r

Explanation

We have
`(x+1) (n) = (1+x)^(n) `
`=.^(n)C_(0) + .^(n)C _(1) x + .^(n) C _ (2) x^(2) + ...+ .^(n) C _(r-2) x ^(r-2) + .^(n) C _(r-1) x^(r-1)`
`+.^(n) C _(r) x ^(r)....`
`:. T _((r-1)) = T _((r-2) + 1) = .^(n)C(r-2) x ^(r-2)," "` ...(i)
`T_(r) = T _((r-1) + 1) = .^(n) C _(r-1) x ^(r-1)," "` ...(ii)
`T_(r+1)= .^(n)C_(r) x ^(r)." "`...(iii)
It is given that
` " coeff. of " T _(r-1) : "coeff. of " T_(r) : "coeff . of " T_(r+1)=1: 3 : 5`
`rArr .^(n)C _((r-2)) : .^(n)C_((r-1)) : .^(n) C _(r) = 1: 3 : 5 `
`rArr (.^(n) C _(r-2) )/(.^(n)C _(r-1)) = 1/3 and (.^(n) C _(r-1))/(.^(n) C r ) = 3/5`.
But,`(.^(n)C_(r-2))/(.^(n)C_(r-1))=1/3 rArr (n!)/(((r-2)!*)(n-r+2)!)xx ((r-1)!*(n-r+1)!)/(n!) =1/3`
` rArr ((r-1))/((n-r+2))=1/3`
`rArr n-4r = -5 " "` ...(i)
And, `(.^(n) C _(r-1))/(.(n) C _(r)) = 3/5 rArr r/ ((n -r+1)) = 3/5 `
`rArr 3n - 8r = - 3." "` ...(ii)
Solving (i) and (ii) , we get `n= 7 and r=3.`
Hence, `n=7and r=3.