Tính S=1^2+2^2+....+100^2
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\(S=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\\ 2S=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}\\ 2S-S=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\right)\\ S=1-\dfrac{1}{2^{100}}=\dfrac{2^{100}-1}{2^{100}}\)
1/1+2+1/1+2+3+1/1+2+3+4+...+1/1+2+3+...+99 +1/50
=1/(2+1).2:2+1/(3+1).3:2+1/(4+1).4:2+..+1/(99+1).99:2+1/50
=2/2.3+2/3.4+2/4.5+..+2/99.100+1/50
=2(1/2.3+1/3.4+1/4.5+..+1/99.100)+1/50
=2(1/2-1/3+1/3-1/4+1/4-1/5+...+1/99-1/100)+1/50
=2(1/2-1/100)+1/50
=49/50+1/50=1
S = 1+1/2.(1+2)+1/3.(1+2+3)+...+1/100.(1+2+3+...+100)
= 1+1/3.(1+2+3)+1/5.(1+2+3+4+5)+...+1/99(1+2+3+...+99) + 1/2.(1+2)+1/4.(1+2+3+4)+...+1/100.(1+2+3+...+100)
= (1+2+3+...+50)+(3/2+5/2+7/2+...+101/2)
= 1275+1300
= 2575
Tham khảo
S =12 + 22 + 32 +......+ 992 + 1002
= 1 + 2.(1 + 1) + 3.(1 + 2) + ... + 99.(1 + 98) + 100.(99 + 1)
= 1 + 2.1 + 2 + 3.1 + 3.2 +... + 99.1 + 99.98 + 100.99 + 100.1
= (2.1 + 2.3 + ... + 99.99 ) + (1 + 2 + 3 + ... + 99 + 100)
= 333300 + 5050
= 338350
\(S=1^2+2^2+3^2+...+99^2+100^2\)
\(=1.1+2.2+3.3+...+100.100\)
\(=1\left(2-1\right)+2\left(3-1\right)+3\left(4-1\right)+...+100\left(101-1\right)\)
\(=\left[1.2-1+2.3-1.1+3.4-3+1+...+100.101-100.1\right]\)
\(=\left[1.2+2.3+3.4+...+100.101\right]-\left(1+2+3+...+100\right)\)
\(=\dfrac{100.101.102}{3}-\dfrac{100.101}{2}\)
\(\dfrac{100.101.\left(2.100+1\right)}{6}=338350\)
S= 21+22+23+...+2100
2S= 22+23+24+...+2100+2101
2S-S= (22+23+24+...+2100+2101)-(21+22+23+...+2100)
S=2101-21
S=2^1+2^2+2^3+...+2^100
2S=2^2+2^3+2^4+...+2^101
2S-S=2^2+2^3+2^4+...+2^101-2^1-2^2-2^3-...-2^100
S=2^101-2^1
mk cũng đang làm bài này, dễ cực luôn
\(B=\frac{5}{28}+\frac{5}{70}+...+\frac{5}{700}\)
\(B=\frac{5}{3}\left[\frac{3}{4.7}-\frac{3}{7.10}+...+\frac{3}{25.28}\right]\)
\(B=\frac{5}{3}\left[\frac{1}{4}-\frac{1}{28}\right]=\frac{5}{14}\)
Chúc bạn học tốt !