Question

Tính tổng:

B=1+4+4²+...+4²⁰⁰+4²⁰¹

^{Giúp mình nha mọi người mình đang cần gấp}

Answer 1

B = 1 + 4 + 4² +...+ 4²⁰⁰ + 4²⁰¹

=> 4B = 4 + 4² +4³ +...+ 4²⁰¹ + 4²⁰²

=> 4B-B = 4²⁰² - 1

3B = 4²⁰² -1

\(\Rightarrow B=\frac{4^{202}-1}{3}\)

Answer 2

4B = 4 + 4^2 + 4^3 + ... + 4^202

4B - B = ( 4 + 4^2 + 4^3 + ... + 4^202 ) - ( 1 + 4 + 4^2 + ... + 4^201 )

3B      = 4^202 - 1

B         = \(\frac{4^{202}-1}{3}\)

Answer 3

\(B=1+4+4^2+...+4^{201}\)

\(\Rightarrow4B=4+4^2+4^3+...+4^{202}\)

\(\Rightarrow4B-B=\left(4+4^2+4^3+...+2^{202}\right)-\left(1+4+4^2+...+4^{201}\right)\)

\(\Rightarrow3B=4^{202}-1\)

\(\Rightarrow B=\frac{4^{202}-1}{3}\)

Vậy  \(B=\frac{4^{202}-1}{3}\)

_Chúc bạn học tốt_

Answer 4

Ta có:

B = 1 + 4 + 4² + ... + 4²⁰⁰ + 4²⁰¹

=> 4B = 4(1 + 4 + 4² + ... + 4²⁰⁰ + 4²⁰¹)

=> 4B = 4 + 4² + 4³ + ... + 4²⁰¹ + 4²⁰²

=> 4B - B = (4 + 4² + 4³ + ... + 4²⁰¹ + 4²⁰²) - (1 + 4 + 4² + ... + 4²⁰⁰ + 4²⁰¹)

=> 3B = 4²⁰² - 1

=> B = (4²⁰² - 1) : 3

=> B = \(\frac{4^{202}-1}{3}\)

Vậy B = \(\frac{4^{202}-1}{3}\)