If `a_0,a_1,a_2,....` be the coefficients in the expansion of `(1+x+x^2)^n` in ascending powers of x. prove that : `(i) a_0a_1-a_1a_2+a_2a_3-....=0`

Explanation

Let `a_(1) , a _(2) , a_(3) and a_(4) " be the coefficients of the rth," (r+1) th, (r+2) th and (r+3) " th terms in the expansion of " (1+x)^(n).` Then,
`a_(1) = .^(n)C _(r-1), a_(2) = .^(n)C_(r), a_(3) = .^(n)C_(r+1) and a_(4) = .^(n) C_(r+2).`
Now, `a_(1) + a_(2) = (.^(n) C_(r-1)+.^(n)C_(r))=.^(n+1)C_(r),`
`a_(3) + a _(4) = (.^(n) C _(r+1) +.^(n) C _(r) )= .^(n+1) C _ (r+2),`
and `a_(2) + a _(3)= (.^(n) C _(r) + .^(n) C _(r+1) ) = .^(n+1) C _(r+1).`
`:. (a_(1))/((a_(1) + a _(2)))+(a_(3))/((a_(3) + a_(4)))= (.^(n)C_(r-1))/(.^(n+1)C_(r))+(.^(n)C_(r+1))/(.^(n+1)C_(r+2))`
`(.^(n) C _ (r-1) )/(((n+1)/r)* (.^(n) C _ (r-1))+(.^(n)C_(r+1))/(((n+1)/(r+2))*^(n)C_(r+1))[ :' .^(n)C_(r) = (n/r)*.^(n-1)C_(r-1)]`
`=r/((n+1))+((r+2))/((n+1)) = (2 (r+1))/((n+1)).`
And , `(2a_(2))/((a_(2) + a_(3)))=(2 xx .^(n)C_(r))/(.^(n+1) C_(r+1))=(2 xx.^(n)C_(r))/(((n+1)/(r+1))* .^(n)C_(r)) = (2(r+1))/(n+1).`
Hence, `(a_(1))/((a_(1) + a_(2)) )+ (a_(3))/((a_(3)+a_(4)))=(2a_(2))/((a_(2) + a_(3)))`.