*) A=1+6+11+16+21+....+101
Dãy trên có: \(\left(101-1\right):5+1=21\)(số số hạng)
\(\Rightarrow A=\frac{\left(101+1\right)\cdot21}{2}=1071\)
*) Đặt C=\(1^2+2^2+3^2+....+98^2=1\cdot1+2\cdot2+3\cdot3+....+98\cdot98\)
\(\Rightarrow B-C=\left(1\cdot2+2\cdot3+3\cdot4+....+98\cdot99\right)-\left(1\cdot1+2\cdot2+3\cdot3+....+98\cdot98\right)\)
\(=\left(1\cdot2-1\cdot1\right)+\left(2\cdot3-2\cdot2\right)+\left(3\cdot4-3\cdot3\right)+.....+\left(98\cdot99-98\cdot98\right)\)
\(=1\left(2-1\right)+2\left(3-2\right)+3\left(4-3\right)+....+98\left(99-1\right)\)
\(=1\cdot1+2\cdot1+3\cdot1+....+98\cdot1\)
\(=1+2+3+....+98\)
\(=\frac{\left(98+1\right)\cdot98}{2}=4851\)
A = 1 + 6 + 11 + 16 +21 +... + 101
Số chữ số của tổng A là :
( 101 - 1 ) : 5 + 1 = 21 (số)
Tổng A = 1 + 6 + ... + 101 = (101 + 1) . 21 : 2 = 1071
B = 1.2 + 2.3 +3.4 + ... + 98.99
3B = 1.2.3 + 2.3.3 +... + 98.99.3
3B = 1.2.3 + 2.3.(4 - 1) + ... + 98.99.(100 - 97)
3B = 1.2.3 + 2.3.4 - 1.2.3 + ... + 98.99.100 - 97.98.99
3B = 98.99.100
B = 323400
\(C=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{100}}\)
\(3C=3-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{99}}\)
\(4C=3C+C=3+\frac{1}{3^{100}}\)
\(C=\frac{3^{101}+1}{4.3^{100}}\)
c) \(C=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+.......+\frac{1}{3^{100}}\)
\(\Rightarrow3C=3-1+\frac{1}{3}-\frac{1}{3^2}+.........+\frac{1}{3^{99}}\)
\(\Rightarrow3C+C=4C=3+\frac{1}{3^{100}}\)\(\Rightarrow C=\frac{3+\frac{1}{3^{100}}}{4}\)
