tinh a= (1/2^2-1)(1/3^2-1)(1/4^2-1)...(1/100^2-1)
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A = 1 + \(\frac{1}{2}\left(1+2\right)\)+ \(\frac{1}{3}\left(1+2+3\right)\)+ .... + \(\frac{1}{100}\left(1+2+3+...+100\right)\)
A = \(1+\frac{1}{2}\cdot\frac{2.3}{2}+\frac{1}{3}\cdot\frac{3.4}{2}+...+\frac{1}{100}\cdot\frac{100.101}{2}\)
A = \(\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+...+\frac{101}{2}\)
A = \(\frac{2+3+4+...+101}{2}\)
A = \(\frac{\left(101+2\right).100}{2}\div2\)
A = \(5150\div2=2575\)
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\(A=\frac{3}{1}+\frac{3}{\frac{\left(2+1\right).2}{2}}+\frac{3}{\frac{\left(3+1\right).3}{2}}+....+\frac{3}{\frac{\left(100+1\right).100}{2}}\)
\(\Rightarrow A=\frac{3}{1}+\frac{6}{2.3}+\frac{6}{3.4}+...+\frac{6}{100.101}\)
\(\Rightarrow A=\frac{3}{1}+6.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{101}\right)\)
\(\Rightarrow A=\frac{3}{1}+6.\left(\frac{1}{2}-\frac{1}{101}\right)\)
\(\Rightarrow A=\frac{3}{1}+\frac{6.99}{202}=\frac{297}{101}+\frac{3}{1}=\frac{600}{101}\)
kết quả k bik có sai k
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1 - 2 - 3 + 4 + 5 - 6 - 7 + 8+ ... + 1993 - 1994
= ( 1 - 2 - 3 + 4 ) = ( 5 - 6 - 7 + 8 ) + ... + 1993 - 1994
= 0 + 0 + ... + 1993 - 1994
= 0 + ( -1 ) = -1
b) ta có 1^2+2^2+...+n^2 = n(n+1)(2n+1)/6
=>2^2+4^2+...+(2n)^2= 2^2(1^2+2^2+...+n^2)= 2n(n+1)(2n+1)/3
và 1^2+2^2+...+(2n+1)^2=(2n+1)(2n+2)(4n+3)/...
=>1^2+3^2+5^2+...+(2n+1)^2 = (2n+1)(2n+2)(4n+3)/6 - 2n(n+1)(2n+1)/3 = (2n+1)(n+1)(2n+3)/3
=>1^2-2^2+3^2-4^2+..... -(2n)^2+(2n+1)^2 = (2n+1)(n+1)(2n+3)/3 - 2n(n+1)(2n+1)/3 = (n+1)(2n+1)
do đó ta có khi n = 100 thì
1^2-2^2+3^2-4^2.....+99^2-100^2+101^2 = (100+1)*(2*100+1)=201*101
Mình cũng không chắc câu b cho lắm
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\(∘backwin\)
\(a ) ( x + 1 ) + ( x + 2 ) + ( x + 3 ) + ... + ( x + 100 ) = 5750\)
\( ( x + x + x + ... + x ) + ( 1 + 2 + 3 + ... + 100 ) = 5750 \)
\( 100 x + ( 1 + 100 ) ×100 : 2 = 5750\)
\(100 x + 5050 = 5750\)
\( 100 x = 5750 − 5050\)
\(100 x = 700\)
\(x = 700 : 100\)
\(x = 7\)
\(b,\) \(B=\)\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2021^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2020}+2021\)
\( B < 1 -\)\(\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2020}-\dfrac{1}{2021}\)
\(B<1-\)\(\dfrac{1}{2021}\)
\(B<\)\(\dfrac{2020}{2021}\)
\(\dfrac{2020}{2021}< 1\)
\(B<1\)
a) (x+1) +(x+2 ) + ...+(x+100)=5750
= 100x + (1+2+3+...+100) = 5750
=100x + 5050 = 5750
--> 100x = 5750-5050=700
--> x=7