Question

Cho A có 25 số hạng,B có 100 số hạng.Tìm A:B, biết:

Cho A=\(\dfrac{1}{1x101}\)+\(\dfrac{1}{2x102}\)+\(\dfrac{1}{3x103}\)+...+\(\dfrac{1}{25x125}\)

Cho B=\(\dfrac{1}{1x26}\)+\(\dfrac{1}{2x27}\)+\(\dfrac{1}{3x28}\)+...+\(\dfrac{1}{100x125}\)

mik đang cần gấp mn giải nhanh giúp mik với mik tích cho.

Answer 1

Lời giải:

$100\times A=\frac{101-1}{1\times 101}+\frac{102-2}{2\times 102}+\frac{103-3}{3\times 103}+...+\frac{125-25}{25\times 125}$

$=1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+\frac{1}{3}-\frac{1}{103}+....+\frac{1}{25}-\frac{1}{125}$

$=(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25})-(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{125})$

$25\times B=\frac{26-1}{1\times 26}+\frac{27-2}{2\times 27}+\frac{28-3}{3\times 28}+...+\frac{125-100}{100\times 125}$

$=1-\frac{1}{26}+\frac{1}{2}-\frac{1}{27}+\frac{1}{3}-\frac{1}{28}+....+\frac{1}{100}-\frac{1}{125}$

$=(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100})-(\frac{1}{26}+\frac{1}{27}+...+\frac{1}{125})$

$=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}-(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{125})$

$=100\times A$

Vậy: $25\times B=100t\times A$

$A:B=25:100=\frac{1}{4}$

Answer 2

bài khó quá.Ai giải giúp mik đc k