If the coefficients of `a^(r-1),\ a^r a n d\ a^(r+1)` in the binomial expansion of `(1+a)^n` are in A.P., prove that `n^2-on(4r+1)+4r^2-2=0.`

Answer

N/a

Explanation

`T_(r)=^(n)C_(r-1).x^(r-1)`
`T_(r-1)=^(n)C_(r).x^(r)`
`" and " T_(r+2)=^(n)C_(r+1).x^(n+1)`
`"Given that ".^(n)C_(r-1),^(n)C_(r)" and ".^(n)C_(r+1) " are in A.P."`
`:. 2.^(n)C_(r)=^(n)C_(r-1)+^(n)C_(r+1)`
`rArr" "2=(^(n)C_(r-1))/(.^(n)C_(r))+(.^(n)C_(r+1))/(.^(n)C_(r))`
`rArr " "2=(r)/(n-r+1)+(n-r)/(r+1)`
`rArr " "2=(r(r+1)+(n-r)(n-r+1))/((n-r+1)(r+1))`
`rArr 2(n-r+1)(r+1)=r(r+1)+(n-r)(n-r+1)`
`rArr 2nr+2n-2n^(2)+2=r^(2)+r+n^(2)-2nr+r^(2)+n-r`
`rArr" "0=n^(2)-4nr-n+4r^(2)-2`
`rArr n^(2)-n(4r+1)+4r^(2)-2=0` Hence Proved