Question

tính:

B= 3¹⁰⁰-3⁹⁹+3⁹⁸-3⁹⁷+...+3²-3+1

Answer 1

Ta có; B= 3¹⁰⁰-3⁹⁹+3⁹⁸-3⁹⁷+...+3²-3+1

            = (3¹⁰⁰-3⁹⁹) + (3⁹⁸-3⁹⁷) + ... + (3²-3) + 1

            = 3⁹⁹ + 3⁹⁷ + ......... + 3 + 1

=> 3B = 3¹⁰⁰ + 3⁹⁹ + 3⁹⁷ + ......... + 3

3B - B = 3¹⁰⁰ - 1

=> B = \(\frac{3^{100}-1}{2}\)

Answer 2

Ta có; B= 3¹⁰⁰-3⁹⁹+3⁹⁸-3⁹⁷+...+3²-3+1

            = (3¹⁰⁰-3⁹⁹) + (3⁹⁸-3⁹⁷) + ... + (3²-3) + 1

            = 3⁹⁹ + 3⁹⁷ + ......... + 3 + 1

=> 3B = 3¹⁰⁰ + 3⁹⁹ + 3⁹⁷ + ......... + 3

3B - B = 3¹⁰⁰ - 1

=> B = \(\frac{3^{100}-1}{3}\)

Answer 3

B=3¹⁰⁰-3⁹⁹+3⁹⁸-3⁹⁷+...+3²-3+1

3B=3¹⁰¹-3¹⁰⁰+3⁹⁹-3⁹⁸+...+3³-3²+3

3B+B=3¹⁰¹+1

4B=3¹⁰¹+1

B=\(\frac{3^{101}+1}{4}\)

Answer 4

B=\(\frac{3^{101}+1}{4}\)