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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