CHo P(x)=100x100+99x99+98x98+......+2x2+x
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P(-1) = 100.(-1)¹⁰⁰ + 99.(-1)⁹⁹ + 98.(-1)⁹⁸ + ... + 2.(-1)² + 1.(-1)
= 100 - 99 + 98 + ... + 2 - 1
= (100 - 99) + (98 - 97) + ... + (2 - 1)
= 1 + 1 + ... + 1 (50 chữ số 1)
= 50
Lời giải:
$P(1)=100.1^{100}+99.1^{99}+....+2.1^2+1$
$=100+99+98+...+2+1=100(100+1):2=5050$
Ta có :
\(A=\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+.................+\dfrac{1}{99.99}+\dfrac{1}{100.100}\)
Ta thấy :
\(\dfrac{1}{2.2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3.3}< \dfrac{1}{2.3}\)
.............................
\(\dfrac{1}{99.99}< \dfrac{1}{98.99}\)
\(\dfrac{1}{100.100}< \dfrac{1}{99.100}\)
\(\Rightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+..................+\dfrac{1}{98.99}+\dfrac{1}{99.100}\)
\(\Rightarrow A< \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...........+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\)
\(\Rightarrow A< 1-\dfrac{1}{100}=\dfrac{99}{100}\)
\(\Rightarrow A< \dfrac{99}{100}\)
\(A=\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+.....+\dfrac{1}{99.99}+\dfrac{1}{100.100}\)
\(A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.....+\dfrac{1}{98.99}+\dfrac{1}{99.100}\)
\(A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+.....+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\)
\(A< 1-\dfrac{1}{100}\)
\(A< \dfrac{99}{100}\)
\(A< B\)
đặt A=1/5x5 +1/6x6 + 1/7x7 + .....+ 1/100x100
=>A>1/5x6 + 1/6x7 +1/7x8 + .... + 1/100x101
=>A>1/5 - 1/6 + 1/6 - 1/7 + +1/7 - 1/8 + ..... + 1/100 - 1/101
=>A> 1/5 - 1/101
=>A>96/505 > 96/576 = 1/6
=>A>1/6
=>A>B
a>1/5x6+1/6x7+...+1/100x101
=1/5-1/6+1/6-1/7+...+1/100-1/101
=1/5-1/101
=101/505-5/101
=96/101
vì 96/101>1/6 nên a>1/6
Tính
A=1x2x3+2x3x3+3x4x3+4x5x3+....+98x99x3
B=1x2+2x3+3x4+4x5+...+98x99
C=1x1+2x2+3x3+4x4+5x5+...+98x98
A=1.2.3+2.3(4-1)+3.4(5-2)+4.5(6-3)+....+98.99(100-97) "." la dau nhan
A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5+....+98.99.100-97.98.99
A=1.2.3+98.99.100
A= 970206
Ta có : B = 1.2 + 2.3 + 3.4 + ..... + 98.99
=> 3B = 0.1.2 + 1.2.3 - 1.2.3 + ...... + 98.99.100
=> 3B = 98.99.100
=> B = \(\frac{98.99.100}{3}\) = 323400