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Cách khác của bài 1:
B=1.3+2.4+3.5+...+97.99+98.100B=1.3+2.4+3.5+...+97.99+98.100
B=1(2+1)+2(3+1)+....+97(98+1)+98(99+1)B=1(2+1)+2(3+1)+....+97(98+1)+98(99+1)
B=1.2+1+2.3+2+....+97.98+97+98.99+98B=1.2+1+2.3+2+....+97.98+97+98.99+98
B=(1.2+2.3+3.4+....+97.98+98.99)+(1+2+3+...+98)B=(1.2+2.3+3.4+....+97.98+98.99)+(1+2+3+...+98)
B=98.99.1003+98.992B=98.99.1003+98.992
B=323400+4851=328251B=323400+4851=328251
1.3+2.4+3.5+...+98.100=22−1+32−1+...+992−1=12+22+32+...+992−99=99.100.1996−99=3282511.3+2.4+3.5+...+98.100=22−1+32−1+...+992−1=12+22+32+...+992−99=99.100.1996−99=328251
Bài 2: A=1.2.3+2.3.4+...+97.98.99<=>4A=1.2.3.4+2.3.4.4+...+97.98.99.4=1.2.3.(4−0)+2.3.4.(5−1)+...+97.98.99.(100−96)A=1.2.3+2.3.4+...+97.98.99<=>4A=1.2.3.4+2.3.4.4+...+97.98.99.4=1.2.3.(4−0)+2.3.4.(5−1)+...+97.98.99.(100−96)
1.2.3.(4−0)+2.3.4.(5−1)+...+97.98.99.(100−96)=1.2.3.4−0.1.2.3+2.3.4.5−1.2.3.4+...+97.98.99.100−96.96.98.99=97.98.99.1001.2.3.(4−0)+2.3.4.(5−1)+...+97.98.99.(100−96)=1.2.3.4−0.1.2.3+2.3.4.5−1.2.3.4+...+97.98.99.100−96.96.98.99=97.98.99.100
Suy ra A=97.98.99.1004=23527350A=97.98.99.1004=23527350
a) =1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101
=1-1/101
=100/101
b) =(2/1.3+2/3.5+2/5.7+...+2/99.101).2,5
=(1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101).2,5
=(1-1/101).2,5
=100/101.2,5
=250/101
c) =(2/2.4+2/4.6+2/6.8+...+2/2008-2/2010).2
=(1/2-1/4+1/4-1/6+1/6-1/8+...+1/2008-1/2010).2
=(1/2-1/2010).2
=1004/1005
A=1.3+2.4+3.5+....+99.101
A=1.(2+1)+2.(3+1)+.....+99.(100+1)
A=1.2+1+2.3+2+3.4+3+....+99.100+99
A=1.2+2.3+3.4+...+99.100+(1+2+3+4+....+99)
Đặt B=1.2+2.3+.....+99.100
=>3B=1.2.3+2.3.(4-1)+.....+99.100.(101-98)
=>3B=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+......+99.100.101-98.99.100
=>3B=99.100.101
=>B=33.100.101=333300
Đặt C=1+2+3+4..+99
=>C=(1+99).99:2=4950
=>A=333300+4950=338250
A=1.3+2.4+3.5+....+99.101
A=1.(2+1)+2.(3+1)+.....+99.(100+1)
A=1.2+1+2.3+2+3.4+3+....+99.100+99
A=1.2+2.3+3.4+...+99.100+(1+2+3+4+....+99)
Đặt B=1.2+2.3+.....+99.100
=>3B=1.2.3+2.3.(4-1)+.....+99.100.(101-98)
=>3B=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+......+99.100.101-98.99.100
=>3B=99.100.101
=>B=33.100.101=333300
Đặt C=1+2+3+4..+99
=>C=(1+99).99:2=4950
=>A=333300+4950=338250
A = 1.3 + 2.4 + 3.5 + ... + 99.101
A = 1.(2+1) + 2.(3+1) + 3.(4+1) + ... + 99.(100+1)
A = 1.2 + 1 + 2.3 + 2 + 3.4 + 3 + ... + 99.100 + 99
A = ( 1.2 + 2.3 + 3.4 + ... + 99.100 ) + ( 1 + 2 + 3 + ... + 99 )
đặt B = 1.2 + 2.3 + 3.4 + ... + 99.100
3B = 1.2.3 + 2.3.3 + 3.4.3 + ... + 99.100.3
3B
Đề bài của bạn sai ở chỗ 99.101 nha, phải là 99.100
a) A = 1.2 + 2.3 + 3.4 + ... + 99.100
=>3A = 1.2.3 + 2.3.3 + 3.4.3 + ... + 98.99.3 + 99.100.3
=>3A = 1.2(3-0) + 2.3(4-1) + 3.4(5-2) + ... + 98.99(100 - 97) + 99.100(101 - 98)
=>3A = 1.2.3 - 0.1.2. + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 98.99.100 - 97.98.99 + 99.100.101 - 98.99.100
=> 3A = 0.1.2 + 99.100.101 = 99.100.101
=> A = (99.100.101) : 3
S = 1.3 + 2.4 + 3.5 + 4.6 + ..... + 99.101 + 100.102
= 1.(2 + 1) + 2(3 + 1) + 3.(4 + 1) + ......... + 99(100 + 1) + 100.(101 + 1)
= 1.2 + 1 + 2.3 + 1 + 3.4 + 3 + ........ + 99.100 + 99 + 100.101 + 100
= (1.2 + 2.3 + 3.4 + ....... + 100.101 ) + (1 + 2 + 3 + ....... + 100)
Ta có công thức :
\(1.2+2.3+3.4+....+n\left(n+1\right)=\frac{n\left(n+1\right)\left(n+2\right)}{3}\)
\(1+2+3+...+n=\frac{n\left(n+1\right)}{2}\)
Áp dụng vào bài toán ta được :
\(S=\frac{100.101.102}{3}+\frac{100.101}{2}\)
= 343400 + 5050
= 348450
1.3+2.4+3.5+...+99.101
= ( 2 -1 ). ( 2 + 1 ) + ( 3 -1) . ( 3 + 1 ) + . . . + ( 100 - 1 ) . ( 100 + 1 )
= \(2^2\) - 1 + \(3^2\) - 1 + . . . + \(100^2\) - 1
= ( \(2^2\) + \(3^2\)+ \(100^2\) ) - ( 1 + 1 + . . . + 1 )
= ( \(2^2\) + \(3^2\)+. . . + \(100^2\)) - 99
Tk mk nha