Question

Tính tổng

A = 1 + 3² +3⁴ + ..... + 3⁹⁸ + 3¹⁰⁰

Answer 1

\(A=1+3^2+3^4+...+3^{98}+3^{100}\)

\(3^2\cdot A=3^2+3^4+3^6+...+3^{100}+3^{102}\)

\(9A-A=\left(3^2+3^4+3^6+...+3^{100}+3^{102}\right)-\left(1+3^2+3^4+...+3^{98}+3^{100}\right)\)

\(8A=3^{102}-1\)

\(\Rightarrow A=\dfrac{3^{102}-1}{8}\)

Answer 2

A = 1 + 3² + 3⁴ + ..... + 3⁹⁸ + 3¹⁰⁰
3A = 3. ( 1 + 3² + 3⁴ + ..... + 3⁹⁸ + 3¹⁰⁰ )
3A = 3. 1 + 3. 3² + 3. 3⁴ + ..... + 3. 3⁹⁸ + 3. 3¹⁰⁰
3A = 3² + 3³ + 3⁴ + ..... + 3¹⁰⁰ + 3¹⁰¹
3A - A = ( 3² + 3³ + 3⁴ + ..... + 3¹⁰⁰ + 3¹⁰¹ ) - ( 1 + 3² + 3⁴ + ..... + 3⁹⁸ + 3¹⁰⁰ )
2A = 3¹⁰¹ - 1
A = ( 3¹⁰¹ - 1 ) : 2