Let a, b, c, d be any four consecutive coefficients in the binomial expansion of  (1 + x)n,thenaa+b+cc+d−2bb+c is equal to

Suppose a=nCr,b=nCr+1,C=nCr+2 and d=nCr+3 Nowaa+b= nCr nCr+nCr+1= nCr n+1Cr+1=r+1n+1Similarly, bb+c=(r+1)+1n+1 and cc+d=(r+2)+1n+1Thus, aa+b+cc+d−2bb+c=0