Cho

- Hướng dẫn giải

a, $C=1+3^1+3^2+3^3+...+3^{11}$

= $\left(1+3^1+3^2\right)$ + $\left(3^3+3^4+3^5\right)$ +...+ $\left(3^9+3^{10}+3^{11}\right)$

= $\left(1+3^1+3^2\right)$ + $3^3.\left(1+3^1+3^2\right)$ + ... + $3^9\left(1+3^1+3^2\right)$

= $\left(1+3^1+3^2\right)$.$\left(1+3^3+...+3^9\right)$

= 13.$\left(1+3^3+...+3^9\right)$$⋮$13

b, $C=1+3^1+3^2+3^3+...+3^{11}$

= $\left(1+3^1+3^2+3^3\right)$ + $\left(3^4+3^5+3^6+3^7\right)$ + $\left(3^8+3^9+3^{10}+3^{11}\right)$

= $\left(1+3^1+3^2+3^3\right)$ + $3^4\left(1+3^1+3^2+3^3\right)$ + $3^8\left(1+3^1+3^2+3^3\right)$

= $\left(1+3^1+3^2+3^3\right)$.$\left(1+3^4+3^8\right)$

= 40.$\left(1+3^4+3^8\right)$$⋮$40