Vì tổng S có 100 SH

Mà 100 chia hết cho 2

Do đó ta có:

5+5^2+5^3+....+5^99+5^100

=(5+5^2)+(5^3+5^4)+...+(5^99+5^100)

=5.(1+5)+5^3.(1+5)+...+5^99.(1+5)

=5.6+5^3.6+...+5^99.6

=6.(5+5^3+...+5^99)

Vì 6 chia hết cho 6

Nên 6.(5+5^3+...+5^99) cũng chia hết cho 6

Vậy S chia hết cho 6

\(S=5+5^2+5^3+5^4+....+5^{99}+5^{100}\)

\(=\left(5+5^2\right)+\left(5^3+5^4\right)+....+\left(5^{99}+5^{100}\right)\)

\(=\left[5\left(1+5\right)\right]+\left[5^3\left(1+5\right)\right]+....+\left[5^{99}\left(1+5\right)\right]\)

\(=5\cdot6+5^3\cdot6+....+5^{99}\cdot6\)

\(=6\left(5+5^3+....+5^{99}\right)\)

\(\Rightarrow S⋮6\)

Do \(\frac{5}{20}>\frac{5}{21};\frac{5}{21}>\frac{5}{22};\frac{5}{22}>\frac{5}{23};\frac{5}{23}>\frac{5}{24}\)

Mà \(\frac{5}{24}>\frac{5}{25}\)

\(\Rightarrow\frac{5}{20}+\frac{5}{21}+\frac{5}{22}+\frac{5}{23}+\frac{5}{24}>\frac{5}{25}+\frac{5}{25}+\frac{5}{25}+\frac{5}{25}+\frac{5}{25}=5.\frac{5}{25}=1\)

Vậy M > 1

Ai thấy đúng k nha

ta co:

M>(5/24)*5>1

a, A= 2 + 2² + 2³ +...+ 2¹⁰⁰

A= ( 2+2² ) + ( 2³+2⁴ ) +...+( 2⁹⁹ + 2¹⁰⁰ )

A= 6+ 2² ( 2+2²)+ ...+ 2⁹⁸ (2+2²)

A=6+ 2².6+ ...+ 2⁹⁸.6

A= 6.(2²+...+2⁹⁸) chia hết cho 3 ( vì 6 chia hết cho 3)

cậu giảng cho mình được không ?

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

\(2\equiv-1\left(mod3\right)\\ \Rightarrow2^{2016}\equiv1\left(mod3\right)\\ \Rightarrow2^{2016}-1\equiv0\left(mod3\right)\\ \Rightarrow A⋮3\)

\(2^{2016}=\left(2^4\right)^{504}=16^{504}\)

16 chia 5 dư 1 nên 16^504 chia 5 dư 1

=> 16^504-1 chia hết cho 5

hay A chia hết cho 5

\(2^{2016}-1=\left(2^3\right)^{672}-1=8^{672}-1⋮7\)

lý luận TT trg hợp A chia hết cho 5

(3;5;7)=1 = > A chia hết cho 105

2;3;4 TT ạ !!

Bài 2:

M = 1/2.3/4.5/6...99/100

Ta có: \(\frac{a}{b}\) = 1 - \(\frac{b-a}{b}\) (a; b; n ∈ N* và b > a)

\(\frac{a+n}{b+n}\) = 1 - \(\frac{b-a}{b+n}\)

\(\frac{a}{b}\) < \(\frac{a+n}{b+n}\)

Áp dụng công thức trên ta có:

\(\frac12<\frac{1+1}{2+1}=\frac23\)

\(\frac34<\frac{3+1}{4+1}=\frac45\)

\(\frac56\) < \(\frac{5+1}{6+1}\) = \(\frac67\)

............................

\(\frac{99}{100}\) < \(\frac{99+1}{100+1}\) = \(\frac{100}{101}\)

Cộng vế với vế ta có:

M = \(\frac12\).\(\frac34\).\(\frac56\)...\(\frac{99}{100}\) < \(\frac23\).\(\frac45\)..\(\frac{100}{101}\) = N

M < N (đpcm)

b; M.N = \(\frac12\).\(\frac34\).\(\frac56\)...\(\frac{99}{100}\).\(\frac23\).\(\frac45\)..\(\frac{100}{101}\)

M.N = \(\frac{1.3.5\ldots99}{3.5\ldots101}\). \(\frac{2.4.6\ldots100}{2.4.6\ldots100}\)

M.N = 1/100.101

\(C=\frac{5}{4}+\frac{5}{4^2}+\frac{5}{4^3}+...+\frac{5}{4^{99}}\)

\(4C=5+\frac{5}{4}+\frac{5}{4^2}+\frac{5}{4^3}+...+\frac{5}{4^{98}}\)

\(4C-C=\left(5+\frac{5}{4}+...+\frac{5}{4^{98}}\right)-\left(\frac{5}{4}+\frac{5}{4^2}+...+\frac{5}{4^{99}}\right)\)

\(3C=5-\frac{5}{4^{99}}\)

\(C=\frac{5-\frac{5}{4^{99}}}{3}\)

\(C=\frac{5}{3}-\frac{5}{4^{99}.3}< C\)

                                đpcm

dễ ẹc!!!!!!!!