Out of proportion expansion

Algebra Level 3

f ( x ) = ( 1 2 x + 2 x 2 ) 743 ( 2 + 3 x 4 x 2 ) 744 f(x) = ( 1 - 2x + 2x^2 )^{743}(2 + 3x - 4x^2)^{744}

The sum of the coefficients of the polynomial obtained by collection of the like terms after the expansion above is?

1 2974 0 1487

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2 solutions

The function f ( x ) f(x) can be expanded into a polynomial with integer coefficients a n a_n as follows:

f ( x ) = ( 1 2 x + 2 x 2 ) 743 ( 2 + 3 x 4 x 2 ) 744 = a 2974 x 2974 + a 2973 x 2973 + a 2972 x 2972 + . . . + a 2 x 2 + a 1 x + a 0 \begin{aligned} f(x) & = (1-2x+2x^2)^{743} (2 + 3x - 4x^2 )^{744} \\ & = a_{2974}x^{2974} + a_{2973}x^{2973} + a_{2972}x ^{2972} + ... + a_{2}x^{2} + a_1x + a_0 \end{aligned}

We note that f ( 1 ) f(1) gives us the sum of the coefficients.

f ( 1 ) = a 2974 + a 2973 + a 2972 + . . . + a 2 + a 1 + a 0 = ( 1 2 + 2 ) 743 ( 2 + 3 4 ) 744 = 1 \begin{aligned} f(1) & = a_{2974} + a_{2973} + a_{2972} + ... + a_{2} + a_1 + a_0 \\ & = (1-2+2)^{743} (2 + 3 - 4)^{744} \\ & = \boxed{1} \end{aligned}

Manish Dash
Apr 17, 2015

Simply substitute x=1 to find the sum of coefficients

Please elaborate.

Nishant Rai - 6 years, 1 month ago

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If you substitute x=1, then you can find the sum of all coefficients

Manish Dash - 6 years, 1 month ago

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