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A=1+2+22+......+2100
=>2A=2+2223+......+2100+2101
=>2A-A=(2+22+23+....+2101)-(1+2+22+.....+2100)
=>A=2101-1
B=3+32+...+350
2B=32+33+..+351
2B-B=(32+33+......+351)-(3+32+...+350)
B=351-3
\(A=2^0+2^1+2^2+.....+2^{1990}\)
\(2A=2\left(2^0+2^1+2^2+.....+2^{1990}\right)\)
\(2A=2^1+2^2+2^3+.....+2^{1991}\)
\(2A-A=\left(2^1+2^2+2^3+.....+2^{1991}\right)-\left(2^0+2^1+2^2+.....+2^{1990}\right)\)
\(A=2^{1991}-2^0=2^{1991}-1\)
\(B=a^0+a^1+a^2+a^3+.....+a^n\)
\(B.a=a^1+a^2+a^3+a^4+.....+a^{n+1}\)
\(B.a-B=\left(a^1+a^2+a^3+a^4+......+a^{n+1}\right)-\left(a^0+a^1+a^2+a^3+.....+a^n\right)\)
\(B.a=a^{n+1}-1\Leftrightarrow B=\dfrac{a^{n+1}-1}{a}\)
\(C=1+3+3^2+.....+3^{50}\)
\(3C=3\left(1+3+3^2+.....+3^{50}\right)\)
\(3C=3+3^2+3^3+.....+3^{51}\)
\(3C-C=\left(3+3^2+3^3+.....+3^{51}\right)-\left(1+3+3^2+.....+3^{50}\right)\)
\(2C=3^{51}-1\Rightarrow C=\dfrac{3^{51}-1}{2}\)
2B = 2^2 +3^2+4^2 + ....+51^2
2B-B= 2^2+3^2+4^2+....+51^2 - 1^2 +2^2 + 3^2 +....+50^2
B= 51^2-1^2
= 50^2
=2500
2C = 3^2+4^2+......+ 51^2
2C-C= 3^2+4^2+....+51^2-2^2+3^2+.....+50^2
C= 51^2-2^2
C= 49^2
2D=2^2+3^2+4^2+......+ 50^2
2D-D= 2^2+3^2+......+50^2-1^2+2^2+....+49^2
D= 50^2- 1^2
D= 49^2
\(2A=\left(1+2+2^2+...+2^{50}\right).2\)
\(2A=2+2^2+2^3+...+2^{51}\)
\(2A-A=\left(2+2^2+2^3+...+2^{51}\right)\)\(-\left(1+2+2^2+...+2^{50}\right)\)
\(A=2^{51}-1\)
Theo bài \(A+1=2^n\)
mà \(A=2^{51}-1\)
\(\Rightarrow A+1=2^{51}-1+1\)
Vậy \(A+1=2^{51}\)
\(\Leftrightarrow n=51\)
Bài 1
a/
\(A=1.\left(2-1\right)+2\left(3-1\right)+3\left(4-1\right)+...+10\left(11-1\right)=\)
\(=\left(1.2+2.3+3.4+...+10.11\right)-\left(1+2+3+...+10\right)=\)
Đặt \(B=1.2+2.3+3.4+...+10.11\)
\(\Rightarrow3B=1.2.3+2.3.3+3.4.3+...+10.11.3=\)
\(=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+10.11.\left(12-9\right)=\)
\(=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-...-9.10.11+10.11.12=\)
\(=10.11.12\Rightarrow B=\frac{10.11.12}{3}=4.10.11\)
\(\Rightarrow A=B-\left(1+2+3+...+10\right)=4.10.11+\frac{10.\left(1+10\right)}{2}=\)
\(=4.10.11+5.11=11.\left(4.10+5\right)=11.45=495\)
b/
\(B=5^2\left(1+2^2+3^2+...+10^2\right)=25.495=12375\)
Bài 2
Số số hạng của M \(=\frac{2n-1-1}{2}+1=n\)
\(M=\frac{n\left[1+\left(2n-1\right)\right]}{2}=n^2\)là số chính phương