10.
\(J=\dfrac{1\cdot2+2\cdot4+3\cdot6+4\cdot8+5\cdot10}{3\cdot4+6\cdot8+9\cdot12+12\cdot16+15\cdot20}\\ =\dfrac{1\cdot2+2\cdot1\cdot2\cdot2+3\cdot1\cdot3\cdot2+4\cdot1\cdot4\cdot2+5\cdot1\cdot5\cdot2}{3\cdot4+2\cdot3\cdot2\cdot4+3\cdot3\cdot3\cdot4+4\cdot3\cdot4\cdot4+5\cdot3\cdot4\cdot4}\\ =\dfrac{\left(1\cdot2\right)\cdot\left(1+2\cdot2+3\cdot3+4\cdot4+5\cdot5\right)}{\left(3\cdot4\right)\cdot\left(1+2\cdot2+3\cdot3+4\cdot4+5\cdot5\right)}\\ =\dfrac{1\cdot2}{3\cdot4}\\ =\dfrac{1\cdot1}{3\cdot2}\\ =\dfrac{1}{6}\)
11.
\(K=3^0+3^1+3^2+...+3^{100}\\ =1\cdot\left(3^0+3^1+3^2+...+3^{100}\right)\\ =\dfrac{3-1}{2}\cdot\left(3^0+3^1+3^2+...+3^{100}\right)\\ =\dfrac{\left(3-1\right)\cdot\left(3^0+3^1+3^2+...+3^{100}\right)}{2}\\ =\dfrac{3^1-3^0+3^2-3^1+3^3-3^2+...+3^{101}-3^{100}}{2}\\ =\dfrac{3^{100}-3^0}{2}=\dfrac{3^{100}-1}{2}\)
12.
\(L=1-5+5^2-5^3+...+5^{98}-5^{99}\\ =1\cdot\left(1-5+5^2-5^3+...+5^{98}-5^{99}\right)\\ =\dfrac{5+1}{6}\cdot\left(1-5+5^2-5^3+...+5^{98}-5^{99}\right)\\ =\dfrac{\left(5+1\right)\cdot\left(1-5+5^2-5^3+...+5^{98}-5^{99}\right)}{6}\\ =\dfrac{5+1-5^2-5+5^3+5^2-5^4-5^3+...+5^{99}+5^{98}-5^{100}-5^{99}}{6}\\ =\dfrac{1-5^{100}}{6}\)
