Question

Câu hỏi

Cho A = 4^0+4^2+...+4^2016 và B=4^2017:3. Tính B-A.

Giúp tớ nhanh nha các tình yêu, tặng triệu like luôn :))

Answer 1

Lời giải:

\(A=4^0+4^2+...+4^{2014}+4^{2016}\)(1)

\(\Rightarrow 4^2.A=4^2+4^4+...+4^{2016}+4^{2018}\) (2)

Lấy (2) trừ (1):

\((4^2-1)A=(4^2+4^4+...+4^{2016}+4^{2018})-(1+4^2+...+4^{2016})\)

\(\Leftrightarrow 15A=4^{2018}-1\Rightarrow A=\frac{4^{2018}-1}{15}\)

Do đó:

\(B-A=\frac{4^{2017}}{3}-\frac{4^{2018}-1}{15}=\frac{5.4^{2017}-4^{2018}+1}{15}\)

\(\Leftrightarrow B-A=\frac{4^{2017}(5-4)+1}{15}=\frac{4^{2017}+1}{15}\)

Answer 2

Ta có:

A=4^0+4^2+...+4^2016

4A=(4^0+4^2+...+4^2016).4

4A=4^1+4^2+...+4^2017

3A=4A-A=(4^1+4^2+...+4^2017)-(4^0+4^2+...+4^2016)

3A=4^2017-1

A=(4^2017-1):3

Ta lại có:

B-A=(4^2017:3)-[(4^2017-1):3]

B-A=[4^2017-(4^2017-1)]:3

B-A=(4^2017-4^2017+1):3

B-A=1:3

B-A=1/3 (1 phần 3)

Vậy B-A = 1/3

Answer 3

A=1+4^1+..+4^2016(1)

4A=4(A=1+\(4^1+4^2+..+4^{2016}\))

=\(4^1+4^2+..+4^{2016}\)+\(4^{2017}\)(2)

Lấy (2)-(1)\(\Rightarrow3a=4^{2017}-1\Rightarrow A=\left(4^{2017}-1\right):3\)

\(\Rightarrow B-A=4^{2017}:3-\left(4^{2017}-1\right):3\)

=\(\left[4^{2017}-4^{2017}+1\right]:3\)

=1:3=\(\dfrac{1}{3}\)

Answer 4

A = 4⁰ + 4² + .......... + 4²⁰¹⁶

4²A = 4² + 4⁴ + .............. + 4²⁰¹⁸

\(\Rightarrow\) ( 4² - 1 )A = ( 4² + 4⁴ + ........ + 4²⁰¹⁸ ) - ( 4⁰ + 4² + ......... + 4²⁰¹⁶ )

\(\Rightarrow\) 15A = 4² + 4⁴ + ....... + 4²⁰¹⁸ - 4⁰ - 4² - ........... - 4²⁰¹⁶

\(\Rightarrow\) 15A = 4²⁰¹⁸ - 4⁰ = 4²⁰¹⁸ - 1

\(\Rightarrow\) A = \(\dfrac{4^{2018}-1}{15}\)

\(\Rightarrow\) B - A = \(\dfrac{4^{2017}}{3}\) - \(\dfrac{4^{2018}-1}{15}\)

\(\Rightarrow\) B - A = \(\dfrac{5.4^{2017}-4^{2018}+1}{15}\)

\(\Rightarrow\) B - A = \(\dfrac{4^{2017}.\left(5-4\right)+1}{15}\)

\(\Rightarrow\) B - A = \(\dfrac{4^{2017}.1+1}{15}\) = \(\dfrac{4^{2017}+1}{15}\)

Vậy B - A = \(\dfrac{4^{2017}+1}{15}\)