Question

GIÚP TỚ VỚI

Rút gọn tổng: B=1+3+3^2+...+3^102

Answer 1

\(3B=3+3^2+.......+3^{103}\)

\(3B-B=\left(3+3^2+.......+3^{103}\right)-\left(1+3+..........+3^{102}\right)\)

\(\Rightarrow2B=3^{103}-1\)

\(\Rightarrow B=\frac{3^{102}-1}{2}\)

Answer 2

\(B=1+3+3^2+...+3^{102}\)

\(3B=3+3^2+3^3+...+3^{103}\)

\(3B-B=\left(3+3^2+3^3+...+3^{103}\right)-\left(1+3+3^2+..+3^{102}\right)\)

\(2B=3^{103}-1\)

\(B=\frac{3^{103}-1}{2}\)

Answer 3

Ta có B=1+3+3²+...+3¹⁰²

     3B=3+3²+3³+...+3¹⁰³

-

       B=1+3+3²+...+3¹⁰²

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=> 3B-B=3¹⁰³-1

=>2B=3¹⁰³-1

=>B=(3¹⁰³-1):2

Chúc bạn học giỏi nha ^-^