by : Khai triển ${{\left( 1+2x+3{{x}^{2}} \right)}^{10}}={{a}_{0}}+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+…+{{a}_{20}}{{x}^{20}}$
a) Hãy tính hệ số ${{a}_{4}}$
C. ${{a}_{4}}=C_{10}^{0}{{.2}^{4}}$
B. ${{a}_{4}}={{2}^{4}}C_{10}^{4}$
C. ${{a}_{4}}=C_{10}^{0}C_{10}^{4}$
D. ${{a}_{4}}=C_{10}^{0}{{.2}^{4}}C_{10}^{4}$
b) Tính tổng $S={{a}_{1}}+2{{a}_{2}}+4{{a}_{3}}+…+{{2}^{20}}{{a}_{20}}$
C. $S={{17}^{10}}$
B. $S={{15}^{10}}$
C. $S={{17}^{20}}$
D. $S={{7}^{10}}$
Hướng dẫn
Đặt $f(x)={{(1+2x+3{{x}^{2}})}^{10}}=\sum\limits_{k=0}^{10}{C_{10}^{k}{{3}^{k}}{{x}^{2k}}{{(1+2x)}^{10-k}}}$
$=\sum\limits_{k=0}^{10}{C_{10}^{k}{{3}^{k}}{{x}^{2k}}\sum\limits_{i=0}^{10-k}{C_{10-k}^{i}{{2}^{10-k-i}}{{x}^{10-k-i}}}}$
$=\sum\limits_{k=0}^{10}{\sum\limits_{i=0}^{10-k}{C_{10}^{k}C_{10-k}^{i}{{3}^{k}}{{2}^{10-k-i}}{{x}^{10+k-i}}}}$
a) Ta có: ${{a}_{4}}=C_{10}^{0}{{.2}^{4}}C_{10}^{4}+$
b) Ta có $S=f(2)={{17}^{10}}$
