C=1+3/2^3+4/2^4+5/2^5+....+100/2^100
Tính C=1/2-(1/3+2/3)+(1/4+2/4+3/4)-(1/5+2/5+3/5+4/5)+...+(1/100+2/100+...+99/100)
bài 1
A=1*2*3+2*3*4+3*4*5+...+99*100*101
B=1*3*5+3*5*7+...+95*97*99
C=2*4+4*6+..+98*100
D=1*2+3*4+5*6+...+99*100
E=1^2+2^2+3^2+...+100^2
G=1*3+2*4+3*5+4*6+...+99*101+100*102
H=1*2^2+2*3^2+3*4^2+...+99*100^2
I=1*2*3+3*4*5+5*6*7+7*8*9+...+98*99*100
K=1^2+3^2+5^2+...+99^2
A = 1*2*3 + 2*3*4 + 3*4*5 ... + 99*100*101
=> 4A = 1*2*3*4 + 2*3*4*4 + 3*4*5*4 + ... +99*100*101*4
=> 4A = 1*2*3*4 + 2*3*4*(5 - 1) + 3*4*5*( 6 - 2) + ... + 99*100*101*(102 - 98)
=> 4A = 1*2*3*4 + 2*3*4*5 - 1*2*3*4 + 3*4*5*6 - 2*3*4*5 + ... + 99*100*101*102 - 98*99*100*101
=> 4A = 99*100*101*102
=> 4A = 101989800
=> A = 25497450
S=1×2+2×3+3×4+4×5+...........+99×100
3S=1×2×3+2×3×(4-1)+3×4×(5-2)+4×5×(6-3)+............+99×100×(101-98)
3S=1×2×3+2×3×4-1×2×3+3×4×5-2×3×4+4×5×6-3×4×5+.............+99×100×101-98×99×100
3S=99×100×101
Tại sao 3S=99×100×101
Các bạn giải thích hộ mình với!
MÌNH CẢM ƠN MỌI NGƯỜI!
S=1×2+2×3+3×4+4×5+...........+99×100
3S=1×2×3+2×3×(4-1)+3×4×(5-2)+4×5×(6-3)+............+99×100×(101-98)
3S=1×2×3+2×3×4-1×2×3+3×4×5-2×3×4+4×5×6-3×4×5+.............+99×100×101-98×99×100
3S=99×100×101
Tại sao 3S=99×100×101
Các bạn giải thích hộ mình với!
MÌNH CẢM ƠN MỌI NGƯỜI!
Giúp tui làm 4 câu này nha các bn ghi rõ bài giải này ra luôn nha
A=!+3+3^2+3^3+...+3^100
B=1+4+4^2+4^3+...+4^100
C=1+5+5^2+5^4+5^6+...+5^100
D=3^100+3^101+3^102+...3^150
Thu gọn các tổng sau:
a) A = 1 + 3 + 3 mũ 2 + 3 mũ 3 + ... + 3 mũ 100
b) B = 1 + 4 + 4 mũ 2 + 4 mũ 3 + 4 mũ 4 + ... + 4 mũ 100
c) C = 1 + 5 mũ 2 + 5 mũ 3 + 5 mũ 6 + .... + 5 mũ 200
d) D = 3 mũ 100 + 3 mũ 101 + 3 mũ 102 + .... + 3 mũ 150
a) 3A = 3 + 3^2 + 3^3 + 3^4 + ... + 3^100 + 3^ 101
=> 3A - A = (3 + 3^2 + 3^3 + 3^4 + ... + 3^100 + 3^ 101) - (1 + 3 + 3 ^2 + 3 ^ 3 + ... + 3 ^100)
=> 2A = 3^101 - 1 => A = (3^101 - 1)/2
b) 4B = 4 + 4 ^ 2 + 4 ^3 + 4 ^ 4 + ... + 4 ^ 100 + 4^ 101
=> 4B - B = (4 + 4 ^ 2 + 4 ^3 + 4 ^ 4 + ... + 4 ^ 100 + 4^ 101) - (1 + 4 + 4 ^ 2 + 4 ^3 + 4 ^ 4 + ... + 4 ^ 100 )
=> 3B = 4^101 - 1 => B = ( 4^101 - 1)/2
c) xem lại đề ý c xem quy luật như thế nào nhé.
d) 3D = 3^101 + 3^ 102 + 3^ 103 + ... + 36 150 + 3^ 151
=> 3D - D = (3^101 + 3^ 102 + 3^ 103 + ... + 36 150 + 3^ 151) - (3 ^100 + 3 ^ 101 + 3 ^ 102 + .... + 3 ^ 150)
=> 2D = 3^ 151 - 3^100 => D = ( 3^ 151 - 3^100)/2
a) Có A=\(1+3+3^2+3^3+....+3^{100}\)
\(\Rightarrow\)3A =\(3\left(1+3+3^2+3^3+...+3^{100}\right)\)=\(3+3^2+3^3+3^4+...+3^{101}\)
\(\Rightarrow2A=3+3^2+3^3+....+3^{101}-1-3-3^2-3^3-....-3^{100}=3^{101}-1\)\(\Rightarrow A=\dfrac{3^{101}-1}{2}\)
Bài b/c/d : bn cứ lm tương tự.
1. (1+1/2).(1+1/2^2).(1+1/2^3)....(1+1/2^100) < 3
2. 1/(5+1)+2/(5^2+1)+4/(5^4+1)+...+ 1024/(5^1024+1) <1/4
3. 3/(1!+2!+3!)+4/(2!+3!+4!)+...+100/(98!+99!+100!) <1/2
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Lần đầu post, mình quên mất chưa nêu câu hỏi. Nhờ các bạn chứng minh dùm 3 câu trên với, cám ơn nhiều ah!
1.\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{2^2}\right)\left(1+\frac{1}{2^3}\right)+...+\left(1+\frac{1}{2^{100}}\right)\)
Đặt \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)
\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\)
\(\Rightarrow2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)
\(\Rightarrow A=1-\frac{1}{2^{100}}\)
Thấy:\(\frac{1}{2^{100}}>0\Rightarrow1-\frac{1}{2^{100}}< 1\)
\(\Rightarrow A< 1\)
Ta có:\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{2^2}\right)\left(1+\frac{1}{2^3}\right)...\left(1+\frac{1}{2^{100}}\right)=A+100< 1+100=101\)
\(101>\left(1+\frac{1}{2}\right)\left(1+\frac{1}{2^2}\right)\left(1+\frac{1}{2^3}\right)...\left(1+\frac{1}{2^{100}}\right)\ge100\)
\(\Rightarrow\left(1+\frac{1}{2}\right)\left(1+\frac{1}{2^2}\right)\left(1+\frac{1}{2^3}\right)...\left(\frac{1}{2^{100}}\right)>\left(\frac{101}{100}\right)^{100}>3\)
*Cách khác:
\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{2^2}\right)\left(1+\frac{1}{2^3}\right)+...+\left(1+\frac{1}{2^{100}}\right)\)
\(=\frac{2+1}{2}.\frac{2^2+1}{2^2}....\frac{2^{100}+1}{2^{100}}\)
Ta thấy:
\(\frac{2+1}{2}>\frac{2^2+1}{2^2}>....>\frac{2^{100}+1}{2^{100}}\)
\(\Rightarrow\frac{2+1}{2}>\frac{2+1}{2}.\frac{2^2+1}{2^2}....\frac{2^{100}+1}{2^{100}}\)
Mà \(\frac{2+1}{2}< 3\)
\(\Rightarrow\frac{2+1}{2}.\frac{2^2+1}{2^2}....\frac{2^{100}+1}{2^{100}}< 3\)
\(\Rightarrow\left(1+\frac{1}{2}\right)\left(1+\frac{1}{2^2}\right)\left(1+\frac{1}{2^3}\right)+...+\left(1+\frac{1}{2^{100}}\right)< 3\)
A= 2^2 + 2^3 + 2^4 + 2^5 +...+ 2^100
B= 3^2 + 3^4 + 3^6 + ...+ 3^100
C=5^1 + 5^3 + 5^5 + ... + 5^99
Tính TỔNG QUÁT: S= a + a^2 + a^3 + a^4 + ...+ a^n
Chứng minh rằng:
a. \(\frac{1}{3^2}+\frac{2}{3^3}+\frac{3}{3^4}+\frac{4}{3^5}+...+\frac{99}{3^{100}}+\frac{100}{3^{101}}< \frac{1}{4}\)
b.\(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}< \frac{1}{3}\)
c.\(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{1}{16}\)
d. \(\frac{1}{5^2}-\frac{2}{5^3}+\frac{3}{5^4}-\frac{4}{5^5}+...+\frac{99}{5^{100}}-\frac{100}{5^{101}}< \frac{1}{36}\)