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

Answer

N/a

Explanation

In the expansion of `(1+x)^(n)`
`T_(r+1) =^(n)C_(r).x^(r )`
`:.` Coefficient of `(r+1)`th term `=^(n)C_(r)`
Similarly coefficient of rth term `=^(n)C_(r-1)`
and coeficient of `(r-1)`th term `=^(n)C_(r-2)`
According to the problem
`.^(n)C_(r-2):^(n)C_(r-1):^(n)C_(r)=1:3:5`
`:. " " (.^(n)C_(r-2))/(.^(n)C_(r-1))=(1)/(3)`
`rArr " "(|uln)/(|ul(r-2)|ul(n-r+2)).(|ul(r-1)|ul(n-r+1))/(|uln) =(1)/(3)`
`rArr " "((r-1)(|ul(r-2).|ul(n-r+1)))/(|ul(r-2)(n-r+2)|uln-r+1)=(1)/(3)`
`rArr" "(r-1)/(n-r+2)=(1)/(3)`
`rArr " "n-r+2=3r-3`
`rArr " "n=4r -5....(1)`
`" and " " "(.^(n)C_(r-1))/(.^(n)C_(r))=(3)/(5)`
`rArr " "(|uln)/(|ul(r-1)|ul(n-r+1)).(|ulr|ul(n-r))/(|uln)=(3)/(5)`
`rArr " "(r|ul(r-1)|ul(n-r))/(|ul(r-1)(n-r+1)|ul(n-r))=(3)/(5)`
`rArr " "(r)/(n-r+1)=(3)/(5)`
`rArr " "3n-3r+3=5r`
`rArr " "(4r-5)-3r+3=5r`[from equation (1)]
`rArr " "4r=12`
`rArr " "r=3`
`" Put " r=3` in equations (1).
`n=4xx3-5=7`
`:. " "n=7 " and " r=3`