how that the coefficient of (r+1) th in the expansion of `(1+x)^(n+1)` is equal to the sum of the coefficients of the r th and (r+1) th term in the expansion of `(1+x)^n`

Explanation

We know that
`(1+x) ^(n) = .^(n) C _(0) + . ^(n) C _(1) x + .^(n) C _ (2) x^(2) + ... + .^(n) C _ (r-1) x ^(r-1) + .^(n) C _(r) x^(r)`
`+ .^(n)C _(r+1) x ^(r+1) + ..." " .`
So the coefficients of `x^(r-1) , x^(r) and x^(r+1) " in the given expansion are " .^(n) C _ (r-1), .^(n)C _(r) , and .^(n) C _(r+1)` respectively.
It is given that `.^(n) C _(r-1), .^(n)C_(r), and .^(n) C _(r+1) ` are in AP.
`:. 2 xx .^(n) C _(r) = .^(n) C _(r-1) + .^(n) C _(r+1)`
` rArr (.^(n) C _(r-1))/(.^(n) C _(r)) + (.^(n) C _(r+1) )/(.^(n) C _(r))= 2 .`
`rArr (n!)/((r-1)!*(n-r+1) !) xx ((r! ) * ( n-r) !)/(n!) `
`+ (n!)/((r+1) !* ( n-r+1)!) xx ((r !) * ( n- r)!)/(n!) = 2 `
`rArr r/((n-r+1))+ ((n-r))/((r+1)) = 2`
`[ :' (n-r+1) ! + ( n - R + 1 ) * { ( n - r) ! } and r ! = r * { ( r - 1 ) ! } ]`
`rArr r (r+1) + ( n- r) ( n- r + 1 ) = 2 ( r+ 1 ) ( n- r + 1 ) `
Hence, `rArr n ^ (2) - n ( 4r + 1 ) + 4r^(2) - 2 =0 .`
`n^(2) - n ( 4r+ 1) + 4r^(2) - 2 = 0 .`