Question
Mathematics Question on Binomial theorem
Coefficient of x11 in the expansion of (1+x2)4(1+x3)7(1+x4)12
1051
1106
1113
1120
1113
Solution
The correct answer is C:1113
The expression can be rewritten as a sum of terms involving various coefficients:
[(1+x2)4⋅(1+x3)7⋅(1+x4)12]
=(4C0+4C1x2+4C2x4+4C3x6+4C4x8)⋅(7C0+7C1x3+7C2x6+7C3x9+…+7C7x21)]
[⋅(12C0+12C1x4+12C2x8+…+12C12x48)]
The objective is to find the term that contains (x11) which corresponds to a power of ( x ) equal to 11. This can be achieved by selecting appropriate combinations of coefficients from each part=(4C0⋅7C1⋅12C2)+(4C1⋅7C3⋅12C0)+(4C2⋅7C1⋅12C1)+(4C4⋅7C1⋅1)]
Calculating these products of coefficients:
=[(1⋅7⋅66)+(4⋅35⋅1)+(6⋅7⋅12)+(1⋅7)=462+140+504+7=1113]
So, the coefficient of (x11) in the given expression is 1113.
In a more natural language:
The expression is composed of three parts, each raised to a certain power, and it's required to determine the coefficient of (x11) in the expanded
form. This involves selecting appropriate coefficients from each part. After calculating the necessary products of coefficients, the coefficient of (x11)is found to be 1113.