wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

If a1,a2,a3,a4 be the coefficeints of four consecutive terms in the expansion of (1+x)n then prove that
a1a1+a2+a3(a3+a4)=2a2(a2+a3).

Open in App
Solution

Let a1,a2,a3 and a4 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,
a1=nCr1,a2=nr,a3=NCr+1 and a4=nCr+2.Now,a1+a2=(nCr1+nCr)=n+1Cr,a3+a4=(nCr+1+nCr+2)=n+1Cr+2,and a2+a3=(nCr+nCr+1)=n+1Cr+1a1(a1+a2)+a3(a3+a4)=nCr1n+1Cr+nCr+1n+1Cr+2=nCr1(n+1r.nCr1)+nCr+1(n+1r+2).nCr+1=r(n+1)+(r+2)(n+1)=2(r+1)(n+1)and,2a2(a2+a3)=2×nCrn+1Cr+1=2×nCr(n+1r+1).nCr=2(r+1)(n+1)hence,a1a1+a2+a3(a3+a4)=2a2(a2+a3)


flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
General Term
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon