if a1,a2,a3a,4 are the coefficients of any four consecutive terms in the expansion of (1+x)n, then a1a1+a2+a3a3+a4 is equal to

Let a1,a2, a3, a4 be respectively the coefficients of(r + 1)th, (r + 21)th, (r + 3)th and (r + 4)th terms in theexpansion of (1 + ,x)n .Then,a1=nCr,   a2=nCr+1,  a3=nCr+2,  a4=′Cr+3Now a1a1+a2+a3a3+a4= nCr nCr+nCr+1+ nCr+2 nCr+2+nCr+3                                     = nCr n+1Cr+1+ nCr+2n+1Cr+3                         ∵nCr+nCr+1=n+1Cr+1= nCrn+1r+1nCr+ nCr+2n+1r+3nCr+2            ∵nCr=nrn−1Cr−1=r+1n+1+r+3n+1=2(r+2)n+1=2 nCr+1 n+1Cr+2=2 nCr+1 nCr+1+nCr+2=2a2a2+a3