Question

$\sum _{0\le i\le }^{}\sum _{j\le 10}^{}{}^{10}{C_j}^j{C}_i$ is equal to

• A. $3^{10}$
• B. $3^{10}-1$
• C. $2^{10}$
• D. $2^{10}-1$

Answer 1

The correct option is A $3^{10}$

$\sum _{0\le i\le }^{}\sum _{j\le 10}^{}{}^{10}{C}_j{}^j{C}_i$

${}^{10}{C}_0{\times }^0{C}_0{\times }^{10}{C}_0\left({}^1{C}_0{\times }^1{C}_1\right){+}^{10}{C}_2\left({}^2{C}_0{+}^2{C}_1{+}^2{C}_2\right){+}^{10}{C}_3\left({}^3{C}_0{+}^3{C_1+}^3{C}_2{+}^3{C}_3\right)$

$+...{+}^{10}{C}_{10}({}^{10}{C_0+}^{10}{C}_1+...{+}^{10}{C}_0)$

${=}^{10}{C}_0.2^0{+}^{10}{C}_1.2^1{+}^{10}{C}_2.2^2+...{+}^{10}{C}_{10}.2^{10}$

$={(1+2)}^{10}=3^{10}.$