Q. 18

# Find a positive value of m for which the coefficient of x2 in the expansion (1 + x)m is 6.

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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