IPMAT Indore 2022 (SA) - The sum of the coefficients of all the terms in the expansion of (5 x-9)^4 is __________. | PYQs + Solutions | AfterBoards
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IPMAT Indore 2022 (SA) PYQs

IPMAT Indore 2022

Modern Math
>
Binomial Theorem

Medium

The sum of the coefficients of all the terms in the expansion of (5x9)4(5 x-9)^{4} is __________.

Entered answer:

Correct Answer: 256
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(5x9)4=4C0(5x)4(9)0+4C1(5x)3(9)1+4C2(5x)2(9)2+4C3(5x)1(9)3+4C4(5x)0(9)4=(1×625x4×1)+(4×125x3×9)+(6×25x2×81)+(4×5x×729)+(1×1×6561)=625x44500x3+12150x214580x+6561\begin{aligned} &(5 x-9)^{4}={ }^{4} C_{0}(5 x)^{4}(-9)^{0} \\ &+{ }^{4} C_{1}(5 x)^{3}(-9)^{1} \\ &+{ }^{4} C_{2}(5 x)^{2}(-9)^{2} \\ &+{ }^{4} C_{3}(5 x)^{1}(-9)^{3} \\ &+{ }^{4} C_{4}(5 x)^{0}(-9)^{4} \\ &=\left(1 \times 625 x^{4} \times 1\right)+\left(4 \times 125 x^{3} \times-9\right)+\left(6 \times 25 x^{2} \times 81\right) \\ &+(4 \times 5 x \times-729)+(1 \times 1 \times 6561) \\ &=625 x^{4}-4500 x^{3}+12150 x^{2}-14580 x+6561 \end{aligned}
sum of cerficients =625+12150+6561450014580=256=625+12150+6561-4500-14580 =256 \\
256\therefore 256

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