\(a,\)Đặt \(A=1+2+2^2+...+2^{99}+2^{100}\)

\(\Rightarrow2A=2+2^2+...+2^{100}+2^{101}\)

\(\Rightarrow2A-A=\left(2+2^2+2^3+...+2^{101}\right)-\left(1+2+2^2+...2^{100}\right)\)

\(\Rightarrow A=2^{101}-1\)

\(b,\)Đặt \(B=5+5^3+5^5+...+5^{97}+5^{99}\)

\(\Rightarrow5^2B=5^3+5^5+...+5^{99}+5^{101}\)

\(\Rightarrow25B-B=\left(5^3+5^5+...+5^{99}+5^{101}\right)-\left(5+5^3+...+5^{99}\right)\)

\(\Rightarrow24B=5^{101}-5\)

\(\Rightarrow B=\frac{5^{101}-5}{24}\)

bn hâm mộ cùng phim với mink a

Ta có:A=\(1+3+3^2+3^3+...+3^{2012}\)

      3A=\(3\cdot\left(1+3+3^2+3^3+...+3^{2012}\right)\)

      3A=\(3+3^2+3^3+3^4+...+3^{2013}\)

   3A-A=\(\left(3+3^2+3^3+3^4+...+3^{2013}\right)-\left(1+3+3^2+3^3+...+3^{2012}\right)\)

     2A=\(3+3^2+3^3+3^4+...+3^{2013}-1-3-3^2-3^3-...-3^{2012}\)

     2A=\(\left(3-3\right)+\left(3^2-3^2\right)+\left(3^3-3^3\right)+...+\left(3^{2012}-3^{2012}\right)+\left(3^{2013}-1\right)\)

    2A=\(0+0+0+...+0+3^{2013}-1\)

    2A=\(3^{2013}-1\)

     A=\(\frac{3^{2013}-1}{2}\)

    B=\(3^{2013}\div2\)

    B=\(\frac{3^{2013}}{2}\)

    VậyB-A=\(\frac{3^{2013}}{2}-\frac{3^{2013}-1}{2}\)

          \(B-A=\frac{3^{2013}-\left(3^{2013}-1\right)}{2}\)

          \(B-A=\frac{3^{2013}-3^{2013}+1}{2}\)

          \(B-A=\frac{1}{2}=0,5\)

a/ta có:s=(1-3+3²-3³)+.................+(3⁹⁶-3⁹⁷+3⁹⁸-3⁹⁹)

=-20+.....................+3⁹⁶.(-20.(1+...................3⁹⁶))

suy ra s chia het cho -20

b/ 3s=3-3²+3³-3⁴+^{.................+}3⁹⁹-3¹⁰⁰

3s+s=(3-3²+3³-3⁴+..........................+3⁹⁹-3¹⁰⁰ +(1-3+3²-3³)+............+3⁹⁸-3⁹⁹)

4s=1-3¹⁰⁰

s=(1-3¹⁰⁰):4

​vì s chia hết cho -20 suy ra s chia hết cho 4 suy ra 1-3¹⁰⁰ chia hêt cho 4 suy ra 3¹⁰⁰:4 dư 1

nếu đúng thì tíc cho mình 2 cái nhé!

\(A=3+3^2+3^3+...+3^{100}\)

\(3A=3^2+3^3+3^4+...+3^{101}\)

\(3A-A=\left(3^2+3^3+...+3^{101}\right)-\left(3+3^2+...+3^{100}\right)\)

\(2A=3^{101}-3\)

\(A=\left(3^{101}-3\right):2\)

Ta có : \(2A+3=3^{101}\)

\(→n=101\)

~ Ủng hộ nhé ~

a) Ta có:

$S=1-3+3^2-3^3+\cdots+3^{98}-3^{99}$

$=(1-3)+(3^2-3^3)+\cdots+(3^{98}-3^{99})$

$=-2-2\cdot3^2-\cdots-2\cdot3^{98}$

$=-2(1+3^2+3^4+\cdots+3^{98})$

$=-2(1+9+9^2+\cdots+9^{49})$

$=-2\cdot\dfrac{9^{50}-1}{9-1}$

$=-\dfrac{9^{50}-1}{4}$

Muốn $S$ chia hết cho $-20$ thì cần:

$S\vdots20$

Nhưng xét modulo $20$:

$9^2=81\equiv1\pmod{20}$

$9^{50}\equiv1\pmod{20}$

Do đó:

$S=-\dfrac{9^{50}-1}{4}$

Ta có thể tính chính xác modulo $20$ bằng cách xét $9^{50}-1$ modulo $80$:

$9^2=81\equiv1\pmod{80}$

$9^{50}=(9^2)^{25}\equiv1\pmod{80}$

nên $S$ chia hết cho $20$.

Vậy: $S\vdots20$

Mà một số chia hết cho $20$ thì cũng chia hết cho $-20$.
b) Tính $S$

Từ: $4S=1-3^{100}$

Suy ra: $S=\dfrac{1-3^{100}}{4}$

Hay: $S=-\dfrac{3^{100}-1}{4}$

Từ đây: $4S=1-3^{100}$

Vì $4S$ chia hết cho $4$ nên:

$1-3^{100}\vdots4$

$3^{100}\equiv1\pmod4$

Vậy $3^{100}$ chia $4$ dư $1$.

a) \(D=\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{100}}\)

\(\Rightarrow7D=1+\frac{1}{7}+\frac{1}{7^2}+...+\frac{1}{7^{99}}\)

\(\Rightarrow7D-D=\left(1+\frac{1}{7}+\frac{1}{7^2}+...+\frac{1}{7^{99}}\right)-\left(\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{100}}\right)\)

\(\Rightarrow6D=1-\frac{1}{7^{100}}\)

\(\Rightarrow D=\left(1-\frac{1}{7^{100}}\right).\frac{1}{6}\)