Chứng minh
a) N= 1/4^2 + 1/6^2 + 1/8^2 + ... + 1/2n^2 < 4 (n\(\(\varepsilon\)\)N, n \(\(\ge\)\)2)
b)P=2!/3! + 2!/4! + 2!/5! + ... + 2!/n! < 1 (nεN, n ≥3)
GIÚP MÌNH VỚI !!!!!!!!!!!
1. CMR: ∀ n∈\(N^{\cdot}\)
a) \(A=5^n+2.3^{n-1}+1\text{⋮}8\)
b) \(B=3^{n+2}+4^{2n+1}\text{⋮}13\)
c) \(C=6^{2n}+3^{n+2}+3^n\text{⋮}11\)
d) \(D=1^n+2^n+5^n+8^n\text{⋮}8\)
2. \(CMR:\) \(1^{2002}+2^{2002}+...+2002^{2002}\text{⋮}11\)
3. a) cho a,b ∈Z, t/m:\(a^2+b^2\text{⋮}7\). \(CMR:a\text{⋮}7;b\text{⋮}7\)
b) \(CMR:\) Nếu \(a^2+b^2\text{⋮}21\) thì \(a^2+b^2\text{⋮}441\) (a,b ∈Z)
\(1,\)
\(a,\) Với \(n=1\Leftrightarrow5+2\cdot1+1=8⋮8\left(đúng\right)\)
Giả sử \(n=k\left(k\ge1\right)\Leftrightarrow5^k+2\cdot3^{k-1}+1⋮8\)
Với \(n=k+1\)
\(5^n+2\cdot3^{n-1}+1=5^{k+1}+2\cdot3^k+1\\ =5^k\cdot5+2\cdot3^k+1\\ =5^k\cdot2+2\cdot3^k+5^k\cdot3+1\\ =2\left(5^k+3^k\right)+5^k+2\cdot5^{k-1}+1+2\cdot3^{k-1}-2\cdot3^{k-1}\\ =2\left(5^k+3^k\right)+\left(5^k+2\cdot3^{k-1}+1\right)-2\left(3^{k-1}+5^{k-1}\right)\)
Vì \(5^k+3^k⋮\left(5+3\right)=8;5^{k-1}+3^{k-1}⋮\left(5+3\right)=8;5^k+2\cdot3^{k-1}+1⋮8\) nên \(5^{k+1}+2\cdot3^k+1⋮8\)
Theo pp quy nạp ta được đpcm
\(b,\) Với \(n=1\Leftrightarrow3^3+4^3=91⋮13\left(đúng\right)\)
Giả sử \(n=k\left(k\ge1\right)\Leftrightarrow3^{k+2}+4^{2k+1}⋮13\)
Với \(n=k+1\)
\(3^{n+2}+4^{2n+1}=3^{k+3}+4^{2k+3}\\ =3^{k+2}\cdot3+16\cdot4^{2k+1}\\ =3^{k+2}\cdot3+3\cdot4^{2k+1}+13\cdot4^{2k+1}\\ =3\left(3^{k+2}+4^{2k+1}\right)+13\cdot4^{2k+1}\)
Vì \(3^{k+2}+4^{2k+1}⋮13;13\cdot4^{2k+1}⋮13\) nên \(3^{k+3}+4^{2k+3}⋮13\)
Theo pp quy nạp ta được đpcm
\(1,\)
\(c,C=6^{2n}+3^{n+2}+3^n\\ C=36^n+3^n\cdot9+3^n\\ C=\left(36^n-3^n\right)+\left(3^n\cdot9+2\cdot3^n\right)\\ C=\left(36^n-3^n\right)+3^n\cdot11\)
Vì \(36^n-3^n⋮\left(36-3\right)=33⋮11;3^n\cdot11⋮11\) nên \(C⋮11\)
\(d,D=1^n+2^n+5^n+8^n\)
Vì \(1^n+2^n+5^n⋮\left(1+2+5\right)=8;8^n⋮8\) nên \(D⋮8\)
\(2,\)
Ta thấy:\(1+2+...+2002=\left(2002+1\right)\left(2002-1+1\right):2=2003\cdot2002:2⋮11\left(2002⋮11\right)\)
Do đó \(1^{2002}+2^{2002}+...+2002^{2002}⋮1+2+...+2002⋮11\)
CMR :
a) N = 1/4^2 + 1/6^2 + 1/8^2 + ... + 1/(2n)^2 < 1/4 ( n thuộc N ; n lớn hơn hoặc bằng 2 )
b) P = 2!/3! + 2!/4! + 2!/5! + ... + 2!/n! < 1 ( n thuộc N ; n lớn hơn hoặc bằng 3 )
a) \(N=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}\)
\(N=\frac{1}{2^2}.\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}\right)\)
Đặt A = \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}\)
A < \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right).n}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\)
\(=1-\frac{1}{n}< 1\)( vì n \(\ge\)2 )
\(\Rightarrow N=\frac{1}{2^2}.\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}\right)< \frac{1}{2^2}.1=\frac{1}{4}\)
Vậy \(N< \frac{1}{4}\)
b) \(P=\frac{2!}{3!}+\frac{2!}{4!}+\frac{2!}{5!}+...+\frac{2!}{n!}\)
\(P=2!\left(\frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!}+...+\frac{1}{n!}\right)\)
\(P< 2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{\left(n-1\right).n}\right)\)
\(P< 2.\left(\frac{1}{2}-\frac{1}{n}\right)=1-\frac{2}{n}< 1\)
Vậy \(P< 1\)
Cmr:
a)M=1/2^2+1/3^2+1/4^2+...+1/n^2<1 (neN;n>=2)
b)N=1/4^2+1/6^2+1/8^2+...+1/(2n)^2<1/4 (n€N,n>=2)
c)P=2!/3!+2!/4!+2!/5!+...+2!/n!<1 (n€N,n>=3)
a)Cho A= 1/2^2+1/3^2+...+1/n^2.CMR A<1
b)Cho B=1/2^2+1/4^2+1/6^2+...+1/(2n)^2.CMR B<1/2
c)Cho C=3/4+8/9+15/16+...+n^2-1/n^2.CMR C<n-2
CMR
a)A= 3/1.4 + 3/4.7 + 3/7.10 + ... + 3/n(n+3) < 1
b)2/5<B=1/2 mũ 2 +1/3 mũ 2 + 1/4 mũ 2 + . . . + 1/9 mũ 2 +1/10 mũ 2
c)C=1/4 mũ 2 + 1/6 mũ 2 + 1/8 mũ 2 + . . . + 1/(2n) mũ 2 <1/4
những ai thích xem minecraft và blockman go thì hãy xem kênh youtube của mik kênh mik là M.ichibi các bn nhớ sud và chia sẻ cho nhiều người khác nhé
1)tính
A=1^2+3^2+5^2+...+(2n-1)^2
B=1^3+3^3+5^3+...+(2n-1)^3
2)tính
A=1x2x3x4+2x3x4x5+...(n-2)x(n-1)
3)tính
B=1x2x4+2x3x5+...+n(n+1)x(n+3)
4)tính
C=2^2+5^2+8^2+...+(3n-1)^2
5)tính
D=1^4+2^4+3^4+...+n^4
GIÚP MÌNH VỚI MÌNH CẦN GẤP
LÀM ĐƯỢC MÌNH CHO 5 SAO
NHANH LÊN NHÉ
Chứng minh rằng:
a) A=1/2+2/2^2+3/2^3+4/4^4+...+100/3^100<2
b) B=1/3+2/3^2+3/3^3+...+100/3^100<3/4
c) C=1/2^3+1/3^3+1/4^3+...+1/n^3<1/4 (n thuộc N; n> hoặc = 2)
d) D=1/3^3+1/4^3+1/5^3+...+1/n^3<1/12 (n thuộc N; n> hoặc =3)
e) E=2/1*4/3*6/5*...*200/199<20
f) F=3/4+5/56+7/144+...+2n+1/n^2+(n+1)^2 ( n nguyên dương)
g) G=1/2*(1/6+1/24+1/60+...+1/9240)>57/62
h) H=1/31+1/32+1/33+...+1/2048>3
i) I=(1-1/3)*(1-1/6)*(1-1/10)*...*(1-1/253)<2/5
j) J=1/2!+2/3!+3/4!+...+n-1/n!<2
k) K=1/2!+5/3!+11/4!+...+n^2+n-1/(n+1)!<2 (n nguyên dương)
l) 1/6<L=1/5^2+1/6^2+1/7^2+...+1/100^2<1/4
a/(Sửa đề bài) A= 1/2 + 2/22 + 3/23 + 4/24 +..+ 100/2100 => 1/2A = 1/22 + 2/23 + 3/24 +..+ 100/2101 => A - 1/2A = 1/2 + 2/22 +..+ 100/2100 - 1/22 - 2/23 -..- 100/2101 => 1/2A = 1/2 + 1/22 + 1/23 +..+ 1/2100 - 100/2101 Gọi riêng cụm (1/2 + 1/22 +..+ 1/2100) là B => 2B = 1 + 1/2 + 1/22 +..+ 1/299 => 2B-B = B = 1+ 1/2 +1/22 +..+ 1/299 - 1/2 - 1/22 -..- 1/2100 = 1 - 1/2100 => 1/2A = 1 - 1/2100 - 100/2101 Có 1/2A < 1 => A < 2 =>ĐPCM b/ => 1/3C = 1/32 + 2/33 + 3/34 +..+ 100/3101 => C - 1/3C = 2/3C = 1/3 + 2/32 +..+ 100/3100 - 1/32 - 2/33 -..- 100/3101 = 1/3 + 1/32 + 1/33 +..+ 1/3100 - 100/3101 Gọi riêng cụm (1/3 + 1/32 +..+ 1/3100) là D => 3D = 1 + 1/3 +..+ 1/399 => 3D - D = 2D = 1 + 1/3 +..+1/399 - 1/3 -1/32 -..- 1/3100 = 1 - 1/3100 => 2/3C *2 = 4/3C = 1 - 1/3100 - 200/3101 Có 4/3C < 1 => C<3/4 => ĐPCM Tạm thời thế đã, giải tiếp đc con nào mình sẽ gửi sau :)
