Question

b, B = 1\(\frac{1}{1x2}+\frac{1}{2x3}+......+\frac{1}{99x100}\)

c, C = \(\frac{1}{1x2}+\frac{1}{2x3}+......+\frac{1}{n\left(n+1\right)}\)

d, D = 1 + 2 + 3 + ......+ n

Answer 1

\(B=1+\frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{99.100}.\)

\(B=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+........+\frac{1}{99}+\frac{1}{100}\)

\(B=1+1-\frac{1}{100}=2-\frac{1}{100}\)

\(B=\frac{199}{100}\)

\(C=\frac{1}{1.2}+\frac{1}{2.3}+........+\frac{1}{n\left(n+1\right)}\)

\(C=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.......+\frac{1}{n}-\frac{1}{n+1}\)

\(C=1-\frac{1}{n+1}\)

\(C=\frac{n+1-1}{n+1}=\frac{n}{n+1}\)

Answer 2

Áp dụng công thức tình dãy số ta có :

\(D=\frac{\left[\left(n-1\right):1+1\right].\left(n+1\right)}{2}=\frac{n.\left(n+1\right)}{2}\)

Answer 3

cảm ơn chị nhiều lắm ạ !

Answer 4

cảm ơn anh luôn  nhé !