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.`

Explanation

We know that

General term of `(a+b)^n`

`T_(r+1)=^nC_r " a^(n-r).b^r`

For `(1+a)^n`

Putting `a=1` and `b=a`

`T_(r+1)=^nC_r " (1)^(n-r).a^r`

`T_(r+1)=^nC_r " a^(r) " " " " ....(1)`

Hence coefficient of `a^r=^nC_r`


Similarly, Finding coefficient of `a^(r-1)`

Putting `r=r-1` in `(1)`

`T_((r-1)+1)=^nC_(r-1) " a^(r-1)`

`T_(r)=^nC_(r-1) " a^(r-1)`

`therefore` Coefficient of `a^(r-1)=^nC_(r-1)`


Finding coefficient of `a^(r+1)`

Putting `r=r+1` in `(1)`

`T_((r+1)+1)=^nC_(r+1) " a^(r+1)`

`T_(r+2)=^nC_(r+1) " a^(r+1)`

`therefore` Coefficient of `a^(r+1)=^nC_(r+1)`


Given that, coefficients of `a^(r-1),a^r`and `a^(r+1)`are in arithmetic progression

i.e. ` " ^nC_(r-1), ^nC_(r), ^nC_(r+1)` are in AP

Therefore, common difference must be equal

` " ^nC_(r) - ^nC_(r-1)=^nC_(r+1) - ^nC_(r) `

` " ^nC_(r) +^nC_(r) =^nC_(r+1)+ ^nC_(r-1)`

`2^nC_(r) =^nC_(r+1)+ ^nC_(r-1)`

` " ^nC_(r+1)+ ^nC_(r-1)=2^nC_(r)`

`(n!)/((r-1)![n-(r-1)]!)+(n!)/((r+1)![n-(r+1)]!)=2xx(n!)/(r!(n-r)!)`

`n!(1/((r-1)![n-(r-1)]!)+1/((r+1)![n-(r+1)]!))=n!(2/(r!(n-r)!))`

`(n!)/(n!)(1/((r-1)![n-(r-1)]!)+1/((r+1)![n-(r+1)]!))=(2/(r!(n-r)!))`

`1/((r-1)!(n-r+1)(n-r)(n-r+1)!)+1/((r+1)r(r-1)!(n-r-1)!)=2/(r(r-1)!(n-r)(n-r-1)!)`

`1/((r-1)!(n-r-1)!)(1/((n-r+1)(n-r))+1/((r+1)r))=2/((r-1)!(n-r-1)![r(n-r)])`

`((r-1)!(n-r-1)!)/((r-1)!(n-r-1)!)(1/((n-r+1)(n-r))+1/((r+1)r))=2/(r(n-r))`

`1/((n-r+1)(n-r))+1/((r+1)r)=2/(r(n-r))`

`((r+1)r+(n-r)(n-r+1))/((n-r+1)(n-r)(r+1)r)=2/(r(n-r))`

`(r+1)r+(n-r)(n-r+1)=(2(n-r)(n-r+1)(r+1)r)/(r(n-r))`

`(r+1)r+(n-r)(n-r+1)=2(n-r+1)(r+1)`

`r^2+r+n(n-r+1)-r(n-r+1)=2[r(n-r+1)+1(n-r+1)]`

`r^2+r+n^2-nr+n-nr+r^2-r=2(rn-r^2+r+n-r+1)`

`2r^2+n^2-2nr+n=-2r^2+2rn+2n+2`

`2r^2+n^2-2nr+n+2r^2-2rn-2n-2=0`

`4r^2+n^2-4nr-n-2=0`

`n^2-4nr-n+4r^2-2=0`

`n^2-n(4r+1)+4r^2-2=0`

Hence proved