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1 what is the coefficient of x^3 in (1+x+x^2+x^3)^12?

 

2 For what value of n<24 are the coefficients of x14 and x24 equal in the expansion of (1+x)^n?

 Jun 27, 2023
 #1
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The coefficient of x^3 in (1+x+x^2+x^3)^12 is 66.

We can use the Binomial Theorem to expand the expression. The Binomial Theorem states that the expansion of (a+b)^n is given by:

(a+b)^n = nC0a^n + nC1a^(n-1)b + nC2a^(n-2)b^2 + ... + nCb^n

n this case, a=1 and b=x. Therefore, the expansion of (1+x)^12 is given by:

(1+x)^12 = 12C0 + 12C1x + 12C2x^2 + 12C3x^3 + ... + 12C11x^11 + 12C12x^12

The coefficient of x^3 is 12C3, which is equal to 66. Therefore, the coefficient of x^3 in (1+x+x^2+x^3)^12 is 66.

 Jun 27, 2023
 #2
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2. The value of n is 16

 Jun 28, 2023

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