Math, asked by bhaktipatel9793, 1 year ago

What is the sum of the coefficients in the expansion of (3 + 2x)99n?

Answers

Answered by Tringa0
1

Answer:

(3+2)^{99} = 5^{99}

Step-by-step explanation:

Here we need to find sum of  the coefficients in the expression (3+2x)^99.

As we know the Binomial expression of  (3+2x)^99 is


(3+2x)^{99} = ^{99}C_{0} (2x)^{99} + ^{99}C_{1} (3)^{1}(2x)^{98}+.........+ ^{99}C_{99} (3)^99


Now to find the sum of all coefficients in the expression we put x =1. Because that will add all the coefficients in the expression , so the sum will be

(3+2*1)^{99} = ^{99}C_{0} (2*1)^{99} + ^{99}C_{1} (3)^{1}(2*1)^{98}+.........+ ^{99}C_{99} (3)^99


Sum of coefficients (3+2)^{99} = 5^{99}


That is the final answer.


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