If the coefficients of mth,(m+1)th and (m+2)th terms in the expansion (1+x)n are in A.P., then
see full answer
Your Exam Success, Personally Taken Care Of
1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya
a
n2+n(4m+1)+4m2+2=0
b
(n+2m)2=n+2
c
(n−2m)2=n+2
d
n2+4(4m+1)+nm2−2=0
answer is C.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
Given expansion (1+x)n is, We have general term in the expansion (x+a)n (∴ Tr+1= nCr xn−r (a)r be the expansion of (x+a)n) mth term: T(m−1)+1= nCm−1 xn−m+1 (a)m−1 The coefficient ofmth term is nCm−1 (m+1)th term: Tm+1= nCm xn−m (a)m The coefficient of(m+1)th term is nCm (m+2)th term: T(m+1)+1= nCm+1 xn−m−1 (a)m+1 The coefficient of(m+2)th term is nCm+1 The coefficients of mth,(m+1)th and (m+2)th terms in the expansion (1+x)n are in A.P., then(if a,b and c are in A.P then b=a+c2⇒2b=a+c )2. nCm= nCm−1+ nCm+1 (∴nCr=n!(n−r)! r!) ⇒2n!m!(n−m)!=n!(m−1)!(n−m+1)!+n!(m+1)!(n−m−1)! ⇒2n!(m−1)!m(n−m−1)!(n−m)=n!(m−1)!(n−m+1)!(n−m)(n−m+1)+n!m(m+1)(m−1)!(n−m−1)! ⇒2n!m(n−m)=n!(n−m)(n−m+1)+n!m(m+1) ⇒2m(n−m)=m(m+1)+(n−m)(n−m+1)(n−m)(n−m+1)m(m+1) ⇒2(n−m+1)(m+1)=m(m+1)+(n−m)(n−m+1) ⇒2(mn−m2+m+n−m+1)=m2+m+n2−nm+n−mn+m2−m ⇒(n−2m)2=n+2