A=1+1/2x3+1/3X6+1/4X10+...+1/16X136

A=1+3/2+2+5/2+3+...+17/2

A=2/2+3/2+4/2+5/2+6/2+...+17/2

A=2+3+4+5+...+16+17/2

A=(2+17)x16:2/2

A=19x16:2/2

A=304:2/2

A=152/2

A=76

****

=11021

11211 - 1 - 1 - 1 - 2 - 2 - 2 - 2 - 3 - 3 - 3 - 4 - 4 - 4 - 5 - 5 - 5 - 6 - 7 - 7 - 65 - 4 - 3 - 2 - 34 - 5 - 3 - 3 - 4

= 11211 - (1 + 1 + 1 + 2 + 2 + 2 + 2 + 3 + 3 + 3 + 4 + 4 + 4 + 5 + 5 + 5 + 6 + 7 + 7 + 65 + 4 + 3 + 2 + 34 + 5 + 3 + 3 + 4)

= 11211 - 190

= 11021

\(1-\frac{1}{2}+2-\frac{2}{3}+3-\frac{3}{4}+4-\frac{1}{4}-3-\frac{1}{3}-2-\frac{1}{2}-1\)

\(=\left(1-1\right)+\left(2-2\right)+\left(3-3\right)+4-2.\frac{1}{2}-\left(\frac{2}{3}+\frac{1}{3}\right)-\left(\frac{3}{4}+\frac{1}{4}\right)\)

\(=4-1-1-1=1\)

Bài 1:

A = 1 + 3 + 3² + ... + 3¹⁰⁰

=> 3A = 3 + 3² + ... + 3¹⁰¹

=> 2A = 3¹⁰¹ - 1

=> A = \(\frac{3^{101}-1}{2}\)

B = 1 + 4² + 4⁴ + ... + 4¹⁰⁰

=> 8B = 4² + 4⁴ + ... + 4¹⁰²

=> 7B = 4¹⁰² - 1

=> B = \(\frac{4^{102}-1}{7}\)

Bài 2:

a) S₁ = 2² + 4² + ... + 20²

=> S₁ = 2²(1+2²+...+10²)

=> S₁ = 2².385

=> S₁ = 1540

b) S₂ = 100² + 200² + ... + 1000²

=> S₂ = 100²(1+2²+...+10²)

=> S₂ = 100².385

=> S₂ = 3850000

\(=5\cdot4\cdot\dfrac{9}{16}-4\cdot\left(-2\right)\cdot\dfrac{3}{4}+\dfrac{1}{2}\cdot\left(-2\right)-\dfrac{9}{4}\)

\(=5\cdot\dfrac{9}{4}+4\cdot4\cdot\dfrac{3}{4}-1-\dfrac{9}{4}\)

\(=\dfrac{45}{4}-\dfrac{9}{4}+4\cdot3-1=9+12-1=20\)

ta có

A = \(1+\frac{1+2}{2}+\frac{1+2+3}{3}+\frac{1+2+3+4}{4}+......+\frac{1+2+3+\text{4 +....+16}}{16}\)

xét tổng S = 1+2+3+4+5+......+n  = \(\frac{\left(n+1\right)n}{2}\) lấy \(\frac{S}{n}=\frac{\frac{\left(n+1\right)n}{2}}{n}=\frac{n+1}{2}\)

ta có

A=\(1+\frac{\frac{2\left(2+1\right)}{2}}{2}+\frac{\frac{3\left(3+1\right)}{2}}{3}+\frac{\frac{4\left(4+1\right)}{2}}{4}+\frac{\frac{5\left(5+1\right)}{2}}{5}+......+\frac{\frac{16\left(16+1\right)}{2}}{16}\)

A = \(1+\frac{1+2}{2}+\frac{1+3}{2}+\frac{1+4}{2}+\frac{1+5}{2}+......+\frac{1+16}{2}\)

A = \(1+\frac{1+2+1+3+1+\text{4+1+5+1+6+.....+1+16}}{2}\)

A = \(1+\frac{151}{2}\)

A = \(\frac{153}{2}\)

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