tinh tong S=[1+2.3+3.5+....+101.201] + [1/\2+ 2/\2+.....+100/\2]
bai1:tinh tong S=1.3+3.5+5.7+...+99.101
bai2 :tinh tong S=1.4+4.7+7.10+...+2017.2020
bai 3: tinh tong N=2.4+4.6+6.8+..+100.102
bai 4: tinh tóng=2.6+6.10+10.14+14.18+...+42.46+50.54
bai 5:tinh tongB=2^2+4^2+6^2+...+100^2
bai 6:C=1^2+3^2+...+100^2
bai7: biet 1^2+2^2+3^2+...+10^2=385 tinh tong 2^2+4^2+6^2+...+20^2
bai 8: tinh tong s=1^2+2^2+3^2+...+99^2
Bài 1 :
\(S=1.3+3.5+5.7+...+99.101=3+15+35+...9999\)
Ta thấy :
\(3=2^2-1\)
\(15=4^2-1\)
\(35=6^2-1\)
.....
\(9999=100^2-1\)
\(\Rightarrow S=2^2+4^2+...+100^2-\left(1\right).\left(\left(100-2\right):2+1\right)\)
\(\Rightarrow S=\dfrac{100.\left(100+1\right)\left(2.100+1\right)}{6}-51\)
\(\Rightarrow S=\dfrac{100.101.201}{6}-51=338299\)
nhanh len nhé mik đang cần gấp ai lam trước mik tích cho
Bài 6 :
\(C=1^2+2^2+...+100^2=\dfrac{100.\left(100+1\right)\left(2.100+1\right)}{6}=\dfrac{100.101.201}{6}=338350\)
Bài 9 :
\(S=1^2+2^2+3^2+...+99^2=\dfrac{99.\left(99+1\right)\left(2.99+1\right)}{6}=\dfrac{99.100.199}{6}=328350\)
tinh tong:
S=(1+2.5+3.5+...+101+201)+12+22+32+...1002)
tính tổng:
S=(1+2.5+3.5...+101+201)+(12+22+32+...1002)
a.1152 - (374+1152)+(-65+374)
b.13-12+11+10-9+8-7-6+5-4+3+2-1
c. tinh tong S = 1.2 + 2.3 + 3.4 + ..... + 99.100
d. chung minh rang : 1/22+1/32 +1/42 + ..... + 1/1002 < 1
c ) S = 1.2 + 2.3 + 3.4 + .... + 99.100
=> 3S = 1.2.3 + 2.3.3 + 3.4.3 + .... + 99.100.3
=> 3S = 1.2.3 + 2.3.( 4 - 1 ) + 3.4.( 5 - 2 ) + .... + 99.100.( 101 - 98 )
=> 3S = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + .... + 99.100.101 - 98.99.100
=> 3S = ( 1.2.3 - 1.2.3 ) + ( 2.3.4 - 2.3.4 ) + .... + ( 98.99.100 - 98.99.100 ) + 99.100.101
=> 3S = 99.100.101 => S = \(\frac{99.100.101}{3}\)
d ) Ta có \(\frac{1}{2^2}<\frac{1}{2.1}=\frac{1}{1}-\frac{1}{2}\)
\(\frac{1}{3^2}<\frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\)
..........
\(\frac{1}{100^2}<\frac{1}{99.100}=\frac{1}{99}-\frac{1}{100}\)
Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{100^2}<\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\)
\(\Leftrightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{100^2}<\frac{1}{1}-\frac{1}{100}=\frac{99}{100}<1\)
1.Cho tong S=5+8+11+14...
tim so hang thu 100 cua day
2.Tinh tong S=1+2+22+...+210
tinh tong S = 1"2 + 2"2 + 3"2 + ..... +100"2 (co cach giai )
1, tinh tong
a, S = 1+2-3-4+5+6-7-8+...-100+101
b,C= 1.2+2.3+...+99.100
c, 1.2.3+2.3.4+...+98.99.100
2,tim so tu nhien x
a, 2+4+6+... +2x=210
b,x + (x-1)+(x-2)+...+(x-50)=255
c, (x+1)+(x+2)+...+(x+100)=5700
3,
S1 =1+2
S2=3+4+5
S3=6+7+8+9
hay tinh S99= ??
mik nham nhe bai 2
c, (x+1)+(x+2)+...+(x+100)=5750
Tinh tong S = 1×2+2×3+3×4+4×5+•••••+99×100
S=1*2+2*3+3*4+...+99*100
3S=3*(1*2+2*3+3*4+...+99*100)
3S=1*2*3+2*3*3+3*4*3+...+99*100*3
3S=1*2*(3-0)+2*3*(4-1)+3*4*(5-2)+...+99*100*(101-98)
3S=1*2*3-1*2*0+2*3*4-2*3*1+3*4*5-3*4*2+...+99*100*101-99*100*98
3S=(1*2*3-2*3*1)+(2*3*4-3*4*2)+...+(98*99*100-99*100*98)+99*100*101
3S=0+0+...+0+999900
3S=999900
S=999900/3
S=333300
3S = 1.2.3 + 2.3.3 + 3.4.3 +...+99.100.3
=1.2.3 + 2.3.(4-1)+3.4(5-2)+...+99.100(101-98)
=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+99.100.101-98.99.100
= 99.100.101
=999900
Nga Nguyễn ơi, bạn chưa chia cho 3 rồi
bai 1 tinh tong
a)1/5.6+1/6.7+1/7.8+.......+1/99.100
b)2/1.3+2/3.5+2/5.7+.........+2/2013.2015
a)1/5.6+1/6.7+1/7.8+.......+1/99.100
= (1/5-1/6)+(1/6-1/7)+(1/7-1/8)+.....+(1/99-1/100)
= 1/5 - 1/100
= 19/100
b)2/1.3+2/3.5+2/5.7+.........+2/2013.2015
= (1/1-1/3)+(1/3-1/5)+(1/5-1/7)+.....+(1/2013+1/2015)
= 1/1 - 1/2015
= 2014/2015
\(a,\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{99.100}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{5}-\frac{1}{100}=\frac{20}{100}-\frac{1}{100}=\frac{19}{100}\)
\(b,\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2013.2015}\)
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2013}-\frac{1}{2015}\)
\(=\frac{1}{1}-\frac{1}{2015}=\frac{2015}{2015}-\frac{1}{2015}=\frac{2014}{2015}\)
Ta có :\(n^2-n=n.\left(n-1\right)\)
\(\implies\) \(n^2=\left(n-1\right)n+n\)
Áp dụng : với n=2017 thay vào ta có:
\(S=1+1.2+2+2.3+3+...+99.100+100\)
\(S=\left(1.2+2.3+...+99.100\right)+\left(1+2+...+100\right)\)
\(S=\frac{99.100.101}{3}+\frac{101.100}{2}\)
\(S=100.101.\left(\frac{99}{3}+\frac{1}{2}\right)\)
\(S=\frac{100.101.201}{6}\)