Question

1. Cho:

C = 1/3+1/3^2+1/3^3+...+1/3^99

Chứng minh C<1/2

2. Cho: A=3/1^2×2^2+5/2^2×3^2+7/3^2×4^2+...+19/9^2×10^2

Giúp mình với chiều nay 2h30 🕝 e ik hc r

Answer 1

\(\dfrac{1}{3}< \dfrac{1}{1.2}\\ \dfrac{1}{3^2}< \dfrac{1}{2.3}\\ \dfrac{1}{3^3}< \dfrac{1}{3.4}\\ ......\\ \dfrac{1}{3^{99}}< \dfrac{1}{99.100}\)

cộng vế với vế, ta được :\(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+......+\dfrac{1}{3^{99}}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.......+\dfrac{1}{99.100}\\ =\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+......+\dfrac{1}{99}-\dfrac{1}{100}=1-\dfrac{1}{100}< 1\)