If \((1+2x+3x^2)^{10}=a_0+a_1x+a_2x^2+\cdots+a_{20}x^{20}\), find the coefficients \(a_1,\ a_2,\ a_3,\) and \(a_{20}\).

Solution

From \((1+2x+3x^2)^{10}=a_0+a_1x+a_2x^2+\cdots+a_{20}x^{20}\), we need to find the coefficients of \(x,x^2,x^3,\) and \(x^{20}\).

Rewrite: \((1+2x+3x^2)^{10}=[1+x(2+3x)]^{10}\).

Using Newton's formula \((a+b)^n=\sum_{i=0}^{n}C(n,i)a^{n-i}b^i\), we get \([1+x(2+3x)]^{10}=C(10,0)+C(10,1)x(2+3x)+C(10,2)x^2(2+3x)^2+\cdots+C(10,10)x^{10}(2+3x)^{10}\).

Coefficient of \(x\): \(a_1=2C(10,1)=20\).

Coefficient of \(x^2\): \(a_2=2^2C(10,2)=4\times45=180\).

Coefficient of \(x^3\): \(a_3=12C(10,2)+8C(10,3)=12\times45+8\times120=1500\).

Coefficient of \(x^{20}\): \(a_{20}=C(10,10)3^{10}=3^{10}\).

Therefore, \(a_1=20,\ a_2=180,\ a_3=1500,\ a_{20}=3^{10}\).

Part 2

Given \((1+x+2x^2)^{20}=a_0+a_1x+a_2x^2+\cdots+a_{40}x^{40}\).

When \(x=1\), \(a_0+a_1+a_2+\cdots+a_{40}=4^{20}\).

When \(x=-1\), \(a_0-a_1+a_2-a_3+\cdots-a_{39}+a_{40}=2^{20}\).

Adding the two equations, \(2a_0+2a_2+2a_4+\cdots+2a_{40}=4^{20}+2^{20}\).

Hence, \(a_0+a_2+a_4+\cdots+a_{40}=\frac{4^{20}+2^{20}}{2}\).

Since \(a_{40}=2^{20}\),

\[ a_0+a_2+a_4+\cdots+a_{38} =\frac{4^{20}+2^{20}}{2}-2^{20} =\frac{2^{40}+2^{20}}{2}-2^{20} =\frac{2^{40}-2^{20}}{2}. \]

Therefore, \(\boxed{a_0+a_2+a_4+\cdots+a_{38}=\frac{2^{40}-2^{20}}{2}}\).