Submitted by Mohammad ElKammash

$\sum_{i=0}^{10} (1+x)^i (2+x)^{10-i} =\frac{(2+x)^{11}-(1+x)^{11}}{(2+x)-(1+x)}=(2+x)^{11}-(1+x)^{11}$ The coefficients of this expression are of the form $2^{11-k} {11 \choose k } - {11 \choose k}$ . Therefore the coefficient of $x^9$ is $3 {11 \choose 9} = 165$

Let, $2+x=a, 1+x=b$ . Now the given expression is $a^{10}+a^9b+ \ldots+ b^{10}=b^{10}[(\frac{a}{b})^{10}+(\frac{a}{b})^9+ \ldots + 1]= b^{10} \frac{(\frac{a}{b})^{11} -1}{(\frac{a}{b})-1} = a^{11}-b^{11} = (2+x)^{11}-(1+x)^{11}$ . So coefficient of $x^9$ in the given expression is $2^2\times ^{11}C_9-^{11}C_9=\boxed{165}$ .

the given expression can be written as $\frac{(2+x)^{11}-(1+x)^{11}}{(2+x)-(1+x)}=(2+x)^{11}-(1+x)^{11}$ and we use binomial cofficient and we know that the coefficient of x^9 in the expansion of (2+x)^11 is $\binom{11}{9}2^{2}=220$ and we know the coefficient of x^9 in the expansion of (1+x)^11 is $\binom{11}{9}1^{2}=55$ so the coefficient of x^9 in the expansion of (2+x)^11-(1+x)^11 is 220 - 55 = 165

Solution #1: Use Wolframalpha's summation function.

Solution #2: Recursive pattern. Note that if the (1+x) values were one less in the 9 terms from the right, then the $x^9$ coefficients would be easily calculated because the sum of the degrees of the polynomials in the terms would be 9, as desired. For example, take $(2+x)^5(1+x)^5$ . If the number of $1+x$ s was 1 less, we would have $(2+x)^5(1+x)^4$ . To obtain $x^9$ , only x's can be chosen, so the coefficient is 1. Because of this, factor out $1+x$ from the rightmost terms, so the expression becomes: $(2+x)^{10}(1+x)^0+(1+x)((2+x)^9(1+x)^0+(2+x)^8(1+x)^1+...+(2+x)^0(1+x)^9)$ . First note that the coefficient of $x^9$ in $(2+x)^{10}(1+x)^0$ is 20 by using the Binomial Theorem. The coefficient of $x^9$ in $(2+x)^9(1+x)^0+(2+x)^8(1+x)^1+...+(2+x)^0(1+x)^9$ is $1+1+...+1 = 10$ . However, there is a problem. The coefficient of $x^8$ also has to be calculated now because we have to multiply by $1+x$ . Thus, we consider recursion. Define $a_n$ to be the coefficient of $x^n$ in $\sum \limits_{i=0}^{n+1} (2+x)^i(1+x)^{10-i}$ . We want $a_9$ . The recursive formula is $a_n=3(n+1)+a_{n-1}$ . To see this, consider what we have done already with $a_9$ . The left term computes to 20, while the factored out part on the right computes to 10 and $a_8$ (the coefficient of $x^8$ ). The left term computes to 20 because by the Binomial Theorem, we have $10 \choose {9}$ $*2=10*2=20$ . Think carefully, and you should be able to understand that $a_n=(n+1)2 + (n+1) + a_{n-1} = 3(n+1) + a_{n-1}$ . We have that $a_0 = 3$ because the coefficient of $x^0$ in $(2+x)+(1+x) = 3+2x$ is 3. Then the answer is $3+3*2+3*2+...3*10 = 3(1+2+...+10) = 3*55 = \boxed{165}$ .

here see :: sum_(n=0)^10 (2+x)^(10-n) (1+x)^n = 11 x^10+165 x^9+1155 x^8+4950 x^7+14322 x^6+29106 x^5+41910 x^4+42075 x^3+28105 x^2+11253 x+2047

this problem is based on the $T_{r+1}$ term finding. For $x^{9}$ from each term we focus on the multiplication of x in each term to get a power 9. From the first term we have to get $x^{9}$ from $(2+x)^{10}$ . Using the formula for $T_{r+1}$ in $(a+x)^{n}$ as $(n)c(r+1) \times a^{r} \times x^{n-r}$ we get power of x in first term as 20 in second term we collect the coefficients from both the expressions by taking into consideration power 9 and 8 in first brackets and 0 and 1 in second brackets and similarly for all the brackets till the last which gives us the coefficient as 10. Now applying the general summation formula for A.P we get summation from 10 to 20 hence also the answer as $\boxed{165}$