Tính tổng 12+32+52+72+...+20112
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uses crt;
var i,n,s:longint;
begin
clrscr;
readln(n);
s:=0;
for i:=1 to n do
s:=s+sqr((2*i-1));
writeln(s);
readln;
end.
\(A=1^2+2^2+3^2+....+10^2\\ A=1^{ }+\left(1+1\right)\cdot2+3\cdot\left(2+1\right)+.....+10\cdot\left(9+1\right)\\ A=1+2\cdot1+2+3\cdot2+3+....+10\cdot9+10\\ A=\left(1+2+3...+10\right)+\left(1\cdot2+3\cdot2+.....+10\cdot9\right)\)
Gọi 1+2+3+...+10 là P
Số số hạng là: (10 - 1) : 1 +1 = 10 (số)
P = (10+1) . 10 : 2 = 55
P = 55
Gọi \(1\cdot2+2\cdot3+....+9\cdot10\) là C
\(C=1\cdot2+2\cdot3+....+9\cdot10\\ 3\cdot C=1\cdot2\cdot3+2\cdot3\cdot3+....+9\cdot10\cdot3\\ 3\cdot C=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+....+9\cdot10\cdot\left(11-8\right)\\ 3\cdot C=1\cdot2\cdot3+2\cdot3\cdot4-1\cdot2\cdot3+.....+9\cdot10\cdot11-8\cdot9\cdot10\\ 3\cdot C=9\cdot10\cdot11\\ 3\cdot C=990\\ C=330\)
\(=>A=P+C\\ =>A=55+330\\ A=385\)
b)
\(B=5^2+10^2+15^2+...+50^2\\ B=5^2+\left(2\cdot5\right)^2+\left(3\cdot5\right)^2+....+\left(5\cdot10\right)^2\\ B=5^2+2^2\cdot5^2+3^2\cdot5^2+...+5^2\cdot10^2\\ B=5^2\cdot\left(1+2^2+3^2+....+10^2\right)\\ B=25\cdot\left(1+2^2+3^2+....+10^2\right)\)
\(\left(1+2^2+3^2+....+10^2\right)=A\)
\(=>B=25\cdot A\\ B=25\cdot385\\ B=9625\)
12 + 52 + 62 = 1 + 25 + 36 = 62
22 + 32 + 72= 4 + 9 + 49 = 62
Vậy 12 + 52 + 62 = 22 + 32 + 72
20112-(304+2012)+(2013+304)
=20112-304-2012+2013+304
=20112+(-2012+2013)+(-304+304)
=20112+1+0=20113
\(\frac{9^{14}.25^5.8^7}{18^{12}.625^3.24^3}=\frac{\left(3^2\right)^{14}.25^5.\left(2^3\right)^7}{2^{12}.\left(3^2\right)^{12}.\left(25^2\right)^3.\left(2^3\right)^3.3^3}=\)\(\frac{3^{28}.25^5.2^{21}}{2^{12}.2^9.3^{24}.3^3.25^6}=\frac{3^{28}.25^5.2^{21}}{2^{21}.3^{27}.25^6}\)\(=\frac{3}{25}\)