Q. 124.6( 48 Votes )
Find a positive value of m for which the coefficient of x2 in the expansion (1 + x)m is 6.
Answer :
The general term Tr+1 in the binomial expansion is given by Tr+1 = nCr an-r br
Here a = 1, b = x and n = m
Putting the value
Tr+1 = mCr 1m-r xr
= mCr xr
We need coefficient of x2
∴ putting r = 2
T2+1 = mC2 x2
The coefficient of x2 = mC2
Given that coefficient of x2 = mC2 = 6
⇒ 
⇒ 
⇒ m(m-1) = 12
⇒ m2- m - 12 =0
⇒ m2- 4m +3m - 12 =0
⇒ m(m-4) + 3(m-4) = 0
⇒ (m+3)(m - 4)= 0
⇒ m = - 3, 4
we need positive value of m so m = 4
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