tính tổng A=1^3+2^3+3^3+....+n^3
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A = \(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2007}}+\frac{1}{3^{2008}}\)
3A= \(1+\frac{1}{3}+...+\frac{1}{3^{2006}}+\frac{1}{3^{2007}}\)
3A-A= \(1-\frac{1}{3^{2008}}\)
a) 3A=1.2.3 + 2.3.3 + 3.4.3 +... + n.(n+1).3
=1.2.(3-0) + 2.3.(4-1) + ... + n.(n+1).[(n+2)-(n-1)]
=[1.2.3+ 2.3.4 + ...+ (n-1).n.(n+1)+ n.(n+1)(n+2)] - [0.1.2+ 1.2.3 +...+(n-1).n.(n+1)]
=n.(n+1).(n+2)
=>S=[n.(n+1).(n+2)] : 3
Câu 6:
uses crt;
var n,i:integer;
begin
clrscr;
readln(n);
for i:=1 to n do
if n mod i=0 then write(i:4);
readln;
end.
5:
uses crt;
var n,i,dem:integer;
begin
clrscr;
readln(n);
dem:=0;
for i:=0 to n do
if i mod 2=1 then
begin
write(i:4);
dem:=dem+1;
end;
writeln;
writeln(dem);
readln;
end.
1/
\(N=1.\left(2-1\right)+2\left(3-1\right)+3\left(4-1\right)+...+99\left(100-1\right)=\)
\(=\left(1.2+2.3+3.4+...+99.100\right)-\left(1+2+3+...+99\right)=\)
Đặt
\(A=1.2+2.3+3.4+...+99.100\)
\(3A=1.2.3+2.3.3+3.4.3+...+99.100.3=\)
\(=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+99.100.\left(101-98\right)=\)
\(=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-...-98.99.100+99.100.101=\)
\(=99.100.101\Rightarrow A=\dfrac{99.100.101}{3}=33.100.101\)
Đặt
\(B=1+2+3+...+99=\dfrac{99.\left(1+99\right)}{2}=4950\)
\(\Rightarrow N=A-B\)
2/
Số hạng cuối cùng là 10000 hoặc 1000000 mới làm được
\(A=1^2+2^2+3^2+...+100^2\)
Tính như câu 1
3/ Làm như bài 4
4/
\(S=1^2+3^2+5^2+...+99^2=\)
\(=1.\left(3-2\right)+3\left(5-2\right)+5\left(7-2\right)+...+99\left(101-2\right)=\)
\(=\left(1.3+3.5+5.7+...+99.101\right)-2\left(1+3+5+...+99\right)\)
Đặt
\(B=1+3+5+...+99=\dfrac{50.\left(1+99\right)}{2}=2500\)
Đặt
\(A=1.3+3.5+5.7+...+99.101\)
\(6A=1.3.6+3.5.6+3.7.6+...+99.101.6=\)
\(=1.3.\left(5+1\right)+3.5.\left(7-1\right)+5.7.\left(9-3\right)+...+99.101.\left(103-97\right)=\)
\(=1.3+1.3.5-1.3.5+3.5.7-3.5.7+5.7.9-...-97.99.101+99.101.103=\)
\(=3+99.101.103\Rightarrow A=\dfrac{3+99.101.103}{6}\)
\(\Rightarrow S=A-2B\)
Bài 1:
\(N=1^2+2^2+3^3+...+99^2\)
\(N=1.1+2.2+3.3+...+99.99\)
\(N=1.\left(2-1\right)+2.\left(3-1\right)+3.\left(4-1\right)+...+99.\left(100-1\right)\)
\(N=1.2-1+2.3-2+3.4-3+...+99.100-99\)
\(N=\left(1.2+2.3+3.4+...+99.100\right)-\left(1+2+3+...+99\right)\)
Đặt \(\left\{{}\begin{matrix}A=1.2+2.3+3.4+...+99.100\\B=1+2+3+...+99\end{matrix}\right.\)
+) Tính \(A=1.2+2.3+3.4+...+99.100\)
Ta có:
\(3A=1.2.3+2.3.3+3.4.3+...+99.100.3\)
\(3A=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+99.100.\left(101-98\right)\)
\(3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+99.100.101-98.99.100\)
\(3A=99.100.101\)
\(\Rightarrow A=\dfrac{99.100.101}{3}=333300\)
+) Tính \(B=1+2+3+...+99\)
\(B\) có số số hạng là: \(\dfrac{99-1}{1}\) + 1 = 99 (số hạng)
\(\Rightarrow B=\dfrac{\left(99+1\right).99}{2}=4950\)
\(\Rightarrow N=A-B=333300-4950=328350\)
\(\Rightarrow N=328350\)
3A=1.2.3+2.3.(4-1)+3.4.(5-2)+...+(n-1)n[(n+1)-(n-2)]
3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+(n-1)n(n+1)-(n-2)(n-1)n
3A=(n-1)n(n+1)
A=(n-1)n(n+1)/3
Ta có :
\(A=1.2+2.3+3.4+...+\left(n-1\right).n\)
\(\Rightarrow3A=1.2.3+2.3.3+3.4.3+...+\left(n-1\right).n.3\)
\(\Rightarrow3A=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+\left(n-1\right).n.\left[\left(n+1\right)-\left(n-2\right)\right]\)
\(\Rightarrow3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+\left(n-1\right)n.\left(n+1\right)-\left(n-2\right).\left(n-1\right).n\)
\(\Rightarrow3A=\left(n-1\right).n.\left(n+1\right)\)
\(\Rightarrow A=\frac{\left(n-1\right).n.\left(n+1\right)}{3}\)
Vậy \(A=\frac{\left(n-1\right).n.\left(n+1\right)}{3}\)
P/s : Mik ko chắc
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