Question

Tính nhanh tổng sau:

A = 1 + 2 + 2 ^ 2 + 2 ^ 3 + ... + 2 ^ 1000

B = 3 + 3 ^ 2 + 3 ^ 3 + ... + 3 ^ 2015

C = 4 + 4 ^ 2 + 4 ^ 3 + ... + 4 ^ n

Answer 1

a)A=1+2+2²+...+2¹⁰⁰⁰

2A=2(1+2+2²+...+2¹⁰⁰⁰)

2A=2+2²+...+2¹⁰⁰¹

2A-A=(2+2²+...+2¹⁰⁰¹)-(1+2+2²+...+2¹⁰⁰⁰)

A=2¹⁰⁰¹-1

b)B=3+3²+...+3²⁰¹⁵

3B=3(3+3²+...+3²⁰¹⁵)

3B=3²+3³+...+3²⁰¹⁶

3B-B=(3²+3³+...+3²⁰¹⁶)-(3+3²+...+3²⁰¹⁵)

2B=2²⁰¹⁶-3

\(B=\frac{2^{2016}-3}{2}\)

c)C=4+4²+...+4ⁿ

4C=4(4+4²+...+4ⁿ)

4C=4²+4³+...+4ⁿ⁺¹

4C-C=(4²+4³+...+4ⁿ⁺¹)-(4+4²+...+4ⁿ)

3C=4ⁿ⁺¹-4

\(C=\frac{4^{n+1}-4}{3}\)

Answer 2

Ta có: A = 1 + 2 + 2² + ...... + 2¹⁰⁰

=> 2A =  2 + 2²  + 2³ + ...... + 2¹⁰¹

=> 2A - A = 2¹⁰¹ - 1

=> A = 2¹⁰¹ - 1

B = 3 + 3² + 3³ + ...... + 2²⁰¹⁵

=> 3B = 3² + 3³ + 3⁴ + ...... + 2²⁰¹⁶

=> 3B - B = 3²⁰¹⁶ - 3

=> 2B = 3²⁰¹⁶ - 3

=> B = 3²⁰¹⁶ - 3/2

 C = 4 + 4² + 4³ + .... + 4ⁿ

=> 4C = 4² + 4³ + 4⁴ + ..... + 4^{n + 1}

=> 4C - C = 4^{n + 1} - 4 

=> 3C = 4^{n + 1} - 4 

=> C = 4^{n + 1} - 4 / 3

Answer 3

\(A=1+2+2^2+2^3+...+2^{1000}\)

\(\Rightarrow2A=2+2^2+2^3+...+2^{1001}\)

\(\Rightarrow2A-A=\left(2+2^2+2^3+...+2^{1001}\right)-\left(1+2+2^2+2^3+...+2^{1000}\right)\)

\(\Rightarrow A=2^{1001}-1\)

\(B=3+3^2+3^3+...+3^{2015}\)

\(\Rightarrow3B=3^2+3^3+3^4+...+3^{2016}\)

\(\Rightarrow3B-B=\left(3^2+3^3+3^4+...+3^{2016}\right)-\left(3+3^2+3^3+...+3^{2015}\right)\)

\(\Rightarrow2B=3^{2016}-3\)

\(\Rightarrow B=\frac{3^{2016}-3}{2}\)

\(C=4+4^2+4^3+...+4^n\)

\(\Rightarrow4C=4^2+4^3+4^4+...+4^{n+1}\)

\(\Rightarrow4C-C=\left(4^2+4^3+4^4+...+4^{n+1}\right)-\left(4+4^2+4^3+...+4^n\right)\)

\(\Rightarrow3C=4^{n+1}-4\)

\(\Rightarrow C=\frac{4^{n+1}-4}{3}\)