Tính C=1.4+2.5+3.6+4.7+...+1006.1009
Tính S=\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...\)
biết tổng S có 100 số hạng.
Tính tổng: S=1.4+2.5+3.6+4.7+...+n.(n+3)
Tính nhanh:
2 . 31 . 12 + 4 . 6 . 42 + 8 . 27 . 3
tính nhanh
\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}\)
S = 3 - \(\frac{3}{100}\)= \(\frac{300}{100}-\frac{3}{100}=\frac{297}{100}\)
S=3/1.4+3/4.7+3/7.10+.....+3/97.100
S=1/1-1/4+1/4-1/7+1/7-1/10+.....+1/97-1/100
S=1-1/100
S=99/100
Cho S =\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{3}{40.43}+\frac{3}{43.46}\)
Hãy C/M S<1
Ta có\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}+\frac{3}{43.46}\)
= \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\)
= \(1-\frac{1}{46}\)
Vì \(1-\frac{1}{46}< 1\)nên S<1
\(S=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+.......+\frac{3}{43\cdot46}\)
\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+......+\frac{1}{43}-\frac{1}{46}\)
\(S=1-\frac{1}{46}\)
Ta có \(1-\frac{1}{46}< 1\)=> S < 1
\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+......+\frac{3}{40.43}+\frac{3}{43.46}\)
\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+.....+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\)
\(S=1-\frac{1}{46}\)
vì \(1-\frac{1}{46}< 1\Rightarrow S< 1\)
tính S = 1.4+2.5+3.6+4.7+...+n.(n+3)
Ta thấy: 1.4 = 1.(1 + 3)
2.5 = 2.(2 + 3)
3.6 = 3.(3 + 3)
4.7 = 4.(4 + 3)
…….
n(n + 3) = n(n + 1) + 2n
Vậy C = 1.2 + 2.1 + 2.3 + 2.2 + 3.4 + 2.3 + … + n(n + 1) +2n
C = 1.2 + 2 +2.3 + 4 + 3.4 + 6 + … + n(n + 1) + 2n
C = [1.2 +2.3 +3.4 + … + n(n + 1)] + (2 + 4 + 6 + … + 2n)
⇒ 3C = 3.[1.2 +2.3 +3.4 + … + n(n + 1)] + 3.(2 + 4 + 6 + … + 2n)
3C = 1.2.3 + 2.3.3 + 3.4.3 + … + n(n + 1).3 + 3.(2 + 4 + 6 + … + 2n)
3C = n(n + 1)(n + 2) +
⇒ C = + =
Tính tổng: S=1.4+2.5+3.6+4.7+...+n.(n+3)
\(S=1.4+2.5+3.6+4.7+...+n\left(n+3\right)\)
\(S=4+10+18+21+...+n\left(n+3\right)\)
S gồm có :
\(\dfrac{n\left(n+3\right)-4}{4}+1\) ( số hạng )
Tổng S là:
\(S=\left[n\left(n+3\right)+4\right].\left[\dfrac{n\left(n+3\right)-4}{4}+1\right]:2\)
\(S=\left(n^2+3n+4\right)\left[\dfrac{n^2+3n-4}{4}+1\right].\dfrac{1}{2}\)
\(S=\dfrac{n^2+3n+4}{2}.\dfrac{n^2+3n}{4}\)
1 số hạng nha mình đang tìm cách giải thích
tính tổng :
S=1.4+2.5+3.6+4.7+...+n.(n+3)
với n=1,2,3,4,5,
Cho S=\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+.....+\frac{3}{40.43}+\frac{3}{43.46}.\)
Hãy chứng tỏ rằng S<1
\(S=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{43}-\frac{1}{46}\)
\(S=1-\frac{1}{46}\)
Có \(1-\frac{1}{46}< 1\)
\(\Rightarrow S< 1\)
nhan xet:3/1.4=1/1-1/4
3/4.7=1/4-1/7
3/7.10=1/7-1/10
.....................
3/40.43=1/40-1/43
3/43.46=1/43-1/46
S=1/1-1/3+1/3-1/7+1/7-1/10+...+1/40-1/43+1/43-1/46
S=1/1-1/46
S=46/46-1/46
S=45/46<1
vay s<1
S=\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}+\frac{3}{43.46}\)
chứng tỏ rằng S<1
S= 1-1/4+1/4-1/7+1/7-1/10+...+1/40-1/43+|1/43-1/46
S= 1-1/46
S= 45/46<1
vậy S<1
duyệt đi
S= 1- 1/4+1/4-1/7+1/7-1/10+...+1/40-1/43+1/43-1/46
S= 1+ (1/4-1/4)+(1/7-1/7)+...+(1/43-1/43)-1/46
S= 1-1/46= 45/46<1
Suy ra S<1
Cho S=\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+.....+\frac{3}{40.43}+\frac{3}{43.46}\)
Hãy chứng minh S <1