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?
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.