quy luật +3+5+7+9+11+........

bai toan nay kho

bài toán này khó duyệt đi

150 thì phải để tính cái đã

baì nào vậy

A = 1 + 4 + 9 + .... + 2401 + 2500

=> A = 12 + 22 + 32 + 42 + ..... + 492 + 502
=> A = 1.1 + 2.2 + 3.3 + ..... + 49.49 + 50.50

=> A = 1.( 2 - 1 ) + 2.( 3 - 1 ) + 3.( 4 - 1 ) + ..... + 49.( 50 - 1 ) + 50.( 51 - 1 )

=> A = 1.2 - 1 + 2.3 - 2 + 3.4 - 3 + ..... + 49.50 - 49 + 50.51 - 50

=> A = ( 1.2 + 2.3 + 3.4 + .... + 49.50 + 50.51 ) - ( 1 + 2 + 3 + ..... + 49 + 50 )

=> A = \(\frac{50.51.52}{3}-\frac{50.\left(50+1\right)}{2}\)
=> A = 44200 - 1275

=> A = 42925

A=1+4+9+16+25+36+49+64+81+100+121+144+169+196+225+.... dài lắm bạn à, trường hợp thế này ko liệt kê ko tính dc

A=1+4+9+......+2401+2500

A=1²+2²+3²+...........+49²+50²

A=1.(2-1)+2.(3-1)+3.(4-1)+.............+49.(50-1)+50.(51-1)

A=1.2-1+2.3-2+3.4-3+.............+49.50-49+50.51-50

A=(1.2+2.3+3.4+............+49.50+50.51)-(1+2+3+.........+49+50)

A=\(\frac{50.51.52}{3}-\frac{50.\left(50+1\right)}{2}\)

A=44200-1275

A=42925

Lời giải:

$\frac{1}{4}< \frac{1}{1.2}$

$\frac{1}{9}< \frac{1}{2.3}$

$\frac{1}{16}< \frac{1}{3.4}$

....

$\frac{1}{2500}< \frac{1}{49.50}$

Cộng theo vế:

$A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{49}-\frac{1}{50}=1-\frac{1}{50}< 1$

Ta có đpcm.

(1/49-1/9)(1/49-1/16).....(1/49-1/49)...(1/49-1/2401)

(1/49-1/9)(1/49-1/16).....0....(1/49-1/2401)

=0

vì khoảng cách là 7 mà 49 chia hết cho 7 nên có phân số 1/49 trong dãy

Cái này là tính 1*1+2*2+3*3+4*4+......+50*50

=(1-1/3)(1-1/4)(1-1/5)*...*(1-1/50)(1+1/3)(1+1/4)*...*(1+1/50)

=2/3*3/4*...*49/50*4/3*5/4*...*51/50

=2/50*51/3=17*1/25=17/25

\(\left(1-\dfrac{1}{9}\right)\cdot\left(1-\dfrac{1}{16}\right)\cdot\left(1-\dfrac{1}{25}\right)\cdot...\cdot\left(1-\dfrac{1}{2500}\right)\)

\(=\left(\dfrac{9}{9}-\dfrac{1}{9}\right)\cdot\left(\dfrac{16}{16}-\dfrac{1}{16}\right)\cdot...\cdot\left(\dfrac{2500}{2500}-\dfrac{1}{2500}\right)\)

\(=\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot\dfrac{24}{25}\cdot...\cdot\dfrac{2499}{2500}\)

\(=\dfrac{8\cdot15\cdot24\cdot...\cdot2499}{9\cdot16\cdot25\cdot...\cdot2500}\)

\(=\dfrac{\left(2\cdot4\right)\cdot\left(3\cdot5\right)\cdot\left(4\cdot6\right)\cdot....\cdot\left(49\cdot51\right)}{\left(3\cdot3\right)\cdot\left(4\cdot4\right)\cdot\left(5\cdot5\right)\cdot...\cdot\left(50\cdot50\right)}\)

\(=\dfrac{\left(2\cdot3\cdot4\cdot5\cdot...\cdot49\right)\left(4\cdot5\cdot6\cdot...\cdot51\right)}{\left(2\cdot3\cdot4\cdot...\cdot50\right)\left(2\cdot3\cdot4\cdot...\cdot50\right)}\)

\(=\dfrac{1\cdot51}{50\cdot2}\)

\(=\dfrac{51}{100}\)