Đặt \(A=1+7^2+7^3+7^4+...+7^{99}\)

\(\Rightarrow7A=7\left(1+7^2+7^3+7^4+...+7^{99}\right)\)

\(\Rightarrow7A=7+7^3+7^4+7^5+...+7^{100}\)

\(\Rightarrow7A-A=\left(7+7^3+7^4+7^5+...+7^{100}\right)-\left(1+7^2+7^3+7^4+...+7^{99}\right)\)

\(\Rightarrow6A=7^{100}-1\)

\(\Rightarrow A=\frac{7^{100}-1}{6}\)

đặt S = 1 + 7² + 7³ + 7⁴ + .... + 7⁹⁹

=> 7S = 7 + 7³ + 7⁴ + 7⁵ + .... 7¹⁰⁰

=> 7S - S = (7 + 7³ + 7⁴ + 7⁵ + .... + 7¹⁰⁰) - (1 + 7² + 7³ + 7⁴ + .... + 7⁹⁹)

=> 6S = (7 + 7¹⁰⁰) - (1 + 7²) 

=> 6S = (7 - 1) + (7¹⁰⁰ - 7²)

=> 6S = 6 + 7¹⁰⁰ - 7²

=> S = \(\frac{6+7^{100}-7^2}{6}\)

rốt cuộc là ai đúng

=\(-19\frac{1}{4}\)

Đặt A = 1 + 7 + 7²+...+7¹⁰¹

=> 7A = 7 + 7²+7³+...+7¹⁰²

=> 7A-A= 7¹⁰²-1

6A = 7¹⁰²-1

\(\Rightarrow A=\frac{7^{102}-1}{6}\)

=\(\dfrac{5}{11}\left(\dfrac{5}{7}+\dfrac{2}{7}\right)+\dfrac{6}{11}\)

=\(\dfrac{5}{11}\times1+\dfrac{6}{11}\)

=\(\dfrac{11}{11}\)=1

\(\dfrac{5}{11}\cdot\dfrac{5}{7}+\dfrac{5}{11}\cdot\dfrac{2}{7}+\dfrac{6}{11}=\dfrac{5}{11}\cdot\left(\dfrac{5}{7}+\dfrac{2}{7}\right)+\dfrac{6}{11}=\dfrac{5}{11}\cdot1+\dfrac{6}{11}=\dfrac{5}{11}+\dfrac{6}{11}=1\)

P=1+(-3)+5+(-7)+...17+(-19)

P=(-2)+(-2)+...+(-2)

P=(-2). 9,5 ( vì có 9,5 số -2) 

P= -19 

\(B=\frac{7}{15}+\frac{4}{5}-1=\frac{19}{15}-1=\frac{4}{15}\)

\(5-\frac{2}{3}+\frac{1}{5}=\frac{13}{3}+\frac{1}{5}=\frac{68}{15}\)