Question

chứng minh rằng

\(\frac{a}{n\left(n+a\right)}\)=\(\frac{1}{n}\)-\(\frac{1}{n+a}\)

áp dụng tính

A=\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+\(\frac{1}{4.5}\)+.....+\(\frac{1}{99.100}\)

B=\(\frac{5}{1.4}\)+\(\frac{5}{4.7}\)+....+\(\frac{5}{100.103}\)

C=\(\frac{1}{15}\)+\(\frac{1}{35}\)+.....+\(\frac{1}{2499}\)

Answer 1

\(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)

\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)

\(A=\frac{1}{2}-\frac{1}{100}\)

\(A=\frac{49}{100}\)

Answer 2

\(B=\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{100.103}\)

\(B=\frac{5}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\right)\)

\(B=\frac{5}{3}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\right)\)

\(B=\frac{5}{3}.\left(\frac{1}{1}-\frac{1}{103}\right)\)

\(B=\frac{510}{103}\)

Answer 3

\(C=\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2499}\)

\(C=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\right)\)

\(C=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)

\(C=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{51}\right)\)

\(C=\frac{8}{51}\)

Answer 4

Phần a chứng minh đâu ạ

Answer 5

ĐS:49/100

Answer 6

Đáp số là : 49/100 nhé bạn