Số số hạng:
(99 - 1) : 2 + 1 = 50 (số hạng)
Số cặp:
50 : 2 = 25 (cặp)
- 1 + 3 - 5 + 7 - ... - 97 + 99
= (3 - 1) + (7 - 5) + . . . + (99 - 97)
= 2 + 2 + . . . + 2
= 2 . 25
= 50
Chứng minh:
\(\dfrac{1}{\sqrt{1}+\sqrt{3}}+\dfrac{1}{\sqrt{5}+\sqrt{7}}+.....+\dfrac{1}{\sqrt{97}+\sqrt{99}}>\dfrac{9}{4}\)
Tính giá trị của biểu thức:
\(A=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+....+\dfrac{1}{97}+\dfrac{1}{99}}{\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.99}+....+\dfrac{1}{97.3}+\dfrac{1}{99.1}}\)
\(B=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+....+\dfrac{1}{99}+\dfrac{1}{100}}{\dfrac{99}{1}+\dfrac{98}{2}+\dfrac{97}{3}+....+\dfrac{1}{99}}\)
Tính tổng : S\(_1\) = \(1+3^2+5^2+7^2+....+97^2+99^2\)
S\(_2\) =\(2+4^2+6^2+8^2+.....+98^2+100^2\)
S\(_3\) = 1.2.3+2.3.4+3.4.5+....+97.98.99
\(\left(100+\dfrac{99}{2}+\dfrac{98}{3}+\dfrac{97}{4}....+\dfrac{1}{100}\right):\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+....\dfrac{1}{100}\right)-2\)
Chứng minh
\(\frac{1}{3\left(1+\sqrt{2}\right)}+\frac{1}{5\left(\sqrt{2}+\sqrt{3}\right)}+\frac{1}{7\left(\sqrt{3}+\sqrt{4}\right)}+...+\frac{1}{97\left(\sqrt{48}+\sqrt{49}\right)}< \frac{3}{7}\)
Tính: A=99^2-98^2+97^2-96^2+...+3^2-2^2+1A=992−982+972−962+...+32−22+1 ta được kết quả là
Tính các tổng sau:
\(T=\dfrac{1}{1+\sqrt{5}}+\dfrac{1}{\sqrt{5}+\sqrt{9}}+\dfrac{1}{\sqrt{9}+\sqrt{13}+......+\dfrac{1}{\sqrt{2013}+\sqrt{2017}}}\)
\(S=\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+.....+\dfrac{1}{100\sqrt{99}+99\sqrt{100}}\)
tính
\(\frac{10}{1+\sqrt{4}}+\frac{10}{\sqrt{4}+\sqrt{7}}+\frac{10}{\sqrt{7}+\sqrt{10}}+...+\frac{10}{\sqrt{97}+\sqrt{100}}\)
tính
\(\frac{10}{1+\sqrt{4}}+\frac{10}{\sqrt{4}+\sqrt{7}}+\frac{10}{\sqrt{7}+\sqrt{10}}+...+\frac{10}{\sqrt{97}+\sqrt{100}}\)
