यदि `(1+a)^(n)` के प्रसार में `a^(r-1),a^(r)` तथा `a^(r+1)` के गुणांक समांतर श्रेणी में हों तो सिद्ध करें कि `n^(2)-n(4r+1)+4r^(2)-2=0`

Explanation

`(1+a)^(n)` के प्रसार में,
`a^(r-1)` का गुणांक `=""^(n)C_(r-1)` तथा `a^(r)` का गुणांक `=""^(n)C_(r)`
प्रश्न से, `""^(n)C_(r-1),""^(n)C_(r),""^(n)C(r+1)A. P.` में है |
`:." "2.""^(n)C_(r)=""^(n)C_(r-1)+""^(n)C_(r+1)`
`implies" "2(n!)/((n-r)!r!)=(n!)/((n-r+1)!(r-1)!)+(n!)/((r+1)!(n-r-1)!)`
`implies" "(2)/((n-r)(n-r-1)!r(r-1)!)=(1)/((n-r+1)(n-r)(n-r-1)!(r-1)!)+(1)/((n-r-1)!(r+1)r(r-1)!)`
`implies" "(2)/(r(n-r))=(1)/((n-r+1)(n-r))+(1)/((r+1)r)`
`implies" "(2)/(r(n-r))=(r(r+1)+(n-r)(n-r+1))/((n-r+1)(n-r)r(r+1))`
`implies" "2[(n-r+1)(r+1)]=r(r+1)+(n-r)(n-r+1)`
`implies" "2nr-2r^(2)+2n+2=r^(2)+r+n^(2)-2nr+r^(2)+n-r`
`implies" "n^(2)-4nr-n+4r^(2)-2=0`
`implies" "n^(2)-n(4r+1)+4r^(2)-2=0`.