Tính 3 n 3 - 5 n 2 + 1 2 n 3 + 6 n 2 + 4 n + 5 .
A. 3
B. 0
C. 3
D. 3 2
1)tính
A=1^2+3^2+5^2+...+(2n-1)^2
B=1^3+3^3+5^3+...+(2n-1)^3
2)tính
A=1x2x3x4+2x3x4x5+...(n-2)x(n-1)
3)tính
B=1x2x4+2x3x5+...+n(n+1)x(n+3)
4)tính
C=2^2+5^2+8^2+...+(3n-1)^2
5)tính
D=1^4+2^4+3^4+...+n^4
GIÚP MÌNH VỚI MÌNH CẦN GẤP
LÀM ĐƯỢC MÌNH CHO 5 SAO
NHANH LÊN NHÉ
1/1*2 +1/2*3 +1/3*4 + 1/4*5 +...+1/n*(n+1) 3/1*2+3/2*3+3/3*4+3/4*5+...+3/n*(n+1) tính tổng nha các bạn
\(S=\dfrac{1}{1x2}+\dfrac{1}{2x3}+\dfrac{1}{3x4}+\dfrac{1}{4x5}+...\dfrac{1}{nx\left(n+1\right)}\)
\(S=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...\dfrac{1}{n}-\dfrac{1}{n+1}\)
\(S=1-\dfrac{1}{n+1}=\dfrac{n}{n+1}\)
\(T=\dfrac{3}{1x2}+\dfrac{3}{2x3}+\dfrac{3}{3x4}+\dfrac{3}{4x5}+...\dfrac{3}{nx\left(n+1\right)}\)
\(T=3x\left[\dfrac{1}{1x2}+\dfrac{1}{2x3}+\dfrac{1}{3x4}+\dfrac{1}{4x5}+...\dfrac{1}{nx\left(n+1\right)}\right]\)
\(T=3x\left[1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...\dfrac{1}{n}-\dfrac{1}{n+1}\right]\)
\(T=3x\left(1-\dfrac{1}{n+1}\right)=\dfrac{3xn}{n+1}\)
viết chương trình tính tổng
s= 1*2/3*4+2*3/4*5+3*4/5*6+...+n*(n+1)/(n+2)*(n+3)
uses crt;
var s:real;
i,n:integer;
begin
clrscr;
readln(n);
s:=0;
for i:=1 to n do
s:=s+(n*(n+1))/((n+2)*(n+3));
writeln(s:4:2);
readln;
end.
Viết chương trình tính tổng : S1 = 1+3+5+7+...+N S2 = 2+4+6+8+...+N S3 = 1-2+3-4+...+N Viết chương trình tính tích : P1 = 1×3×5×7×...× N P2 = 2×4×6×8×...×N
Câu 2:
#include <bits/stdc++.h>
using namespace std;
double p1,p2;
int i,n;
int main()
{
cin>>n;
p1=1;
p2=1;
for (i=1; i<=n; i++)
{
if (i%2==0) p2=p2*(i*1.0);
else p1=p1*(i*1.0);
}
cout<<fixed<<setprecision(2)<<p1<<endl;
cout<<fixed<<setprecision(2)<<p2;
return 0;
}
Tính Tổng:
a) 1*2+2*3+3*4+................+n(n+1)
b) 1*2*3+2*3*4+3*4*5+.......+n(n+1)(n+2)
GIÚP EM VỚI Ạ
EM ĐANG CẦN GẤP Ạ
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
b)
Nhân 4 vào hai vế ta được:
4A = 4.[1.2.3 + 2.3.4 + 3.4.5 + … + (n – 1).n.(n + 1)]
4A = 1.2.3.4 + 2.3.4.4 + 3.4.5.4 + … + (n – 1).n.(n + 1).4
4A = 1.2.3.4 + 2.3.4.(5 – 1) + 3.4.5.(6 – 2) + … + (n – 1).n.(n + 1).[(n + 2) – (n – 2)]
4A = 1.2.3.4 + 2.3.4.5 – 1.2.3.4 + 3.4.5.6 – 2.3.4.5 + … + (n – 1).n(n + 1).(n + 2) – (n – 2).(n – 1).n.(n + 1)
4A = (n – 1).n(n + 1).(n + 2)
A = (n – 1).n(n + 1).(n + 2) : 4.
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
Cho phép cộng n! = 1 . 2 . 3 ... n. Tính 1 . 1! + 2 . 2! + 3 . 3! + 4 . 4! + 5 . 5! = ...............
Tính tổng :
C= 1/1*2*3*4 + 1/2*3*4*5 + ...+ 1/ n*(n+1)*(n+2)*(n+3)
A=(1/1.2.3-1/2.3.4)+(1/2.3.4-1/3.4.5)+..............+(1/n(n+1)(n+2)-1/(n+1)(n+2)(n+3))
A=1/1.2.3-1/(n+1)(n+2)(n+3)
A=1/18-1/(n+1)(n+2)(n+3)
đúng nhé
1) tính \(\lim\limits_{n\rightarrow\infty}\dfrac{3n^5+3n^3-1}{n^3-2n}\)
2) tính \(\lim\limits_{n\rightarrow\infty}\dfrac{3n^7+3n^5-n}{3n^2-2n}\)
1:
\(\lim\limits_{n\rightarrow\infty}\dfrac{3n^5+3n^3-1}{n^3-2n}=\lim\limits_{n\rightarrow\infty}\dfrac{n^5\left(3+\dfrac{3}{n^2}-\dfrac{1}{n^5}\right)}{n^3\left(1-\dfrac{2}{n^2}\right)}\)
\(=\lim\limits_{n\rightarrow\infty}n^2\cdot3=+\infty\)
2: \(\lim\limits_{n\rightarrow\infty}\dfrac{3n^7+3n^5-n}{3n^2-2n}=\lim\limits_{n\rightarrow\infty}\dfrac{3n^6+3n^4-1}{3n-2}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{n^6\left(3+\dfrac{3}{n^2}-\dfrac{1}{n^6}\right)}{n\left(3-\dfrac{2}{n}\right)}=\lim\limits_{n\rightarrow\infty}n^5=+\infty\)
Tính các giới hạn sau:
a) \(\lim \frac{{2{n^2} + 6n + 1}}{{8{n^2} + 5}}\)
b) \(\lim \frac{{4{n^2} - 3n + 1}}{{ - 3{n^3} + 5{n^2} - 2}}\);
c) \(\lim \frac{{\sqrt {4{n^2} - n + 3} }}{{8n - 5}}\);
d) \(\lim \left( {4 - \frac{{{2^{n + 1}}}}{{{3^n}}}} \right)\)
e) \(\lim \frac{{{{4.5}^n} + {2^{n + 2}}}}{{{{6.5}^n}}}\)
g) \(\lim \frac{{2 + \frac{4}{{{n^3}}}}}{{{6^n}}}\).
a) \(\lim \frac{{2{n^2} + 6n + 1}}{{8{n^2} + 5}} = \lim \frac{{{n^2}\left( {2 + \frac{6}{n} + \frac{1}{{{n^2}}}} \right)}}{{{n^2}\left( {8 + \frac{5}{{{n^2}}}} \right)}} = \lim \frac{{2 + \frac{6}{n} + \frac{1}{n}}}{{8 + \frac{5}{n}}} = \frac{2}{8} = \frac{1}{4}\)
b) \(\lim \frac{{4{n^2} - 3n + 1}}{{ - 3{n^3} + 6{n^2} - 2}} = \lim \frac{{{n^3}\left( {\frac{4}{n} - \frac{3}{{{n^2}}} + \frac{1}{{{n^3}}}} \right)}}{{{n^3}\left( { - 3 + \frac{6}{n} - \frac{2}{{{n^3}}}} \right)}} = \lim \frac{{\frac{4}{n} - \frac{3}{{{n^2}}} + \frac{1}{{{n^3}}}}}{{ - 3 + \frac{6}{n} - \frac{2}{{{n^3}}}}} = \frac{{0 - 0 + 0}}{{ - 3 + 0 - 0}} = 0\).
c) \(\lim \frac{{\sqrt {4{n^2} - n + 3} }}{{8n - 5}} = \lim \frac{{n\sqrt {4 - \frac{1}{n} + \frac{3}{{{n^2}}}} }}{{n\left( {8 - \frac{5}{n}} \right)}} = \frac{{\sqrt {4 - 0 + 0} }}{{8 - 0}} = \frac{2}{8} = \frac{1}{4}\).
d) \(\lim \left( {4 - \frac{{{2^{{\rm{n}} + 1}}}}{{{3^{\rm{n}}}}}} \right) = \lim \left( {4 - 2 \cdot {{\left( {\frac{2}{3}} \right)}^{\rm{n}}}} \right) = 4 - 2.0 = 4\).
e) \(\lim \frac{{{{4.5}^{\rm{n}}} + {2^{{\rm{n}} + 2}}}}{{{{6.5}^{\rm{n}}}}} = \lim \frac{{{{4.5}^{\rm{n}}} + {2^2}{{.2}^{\rm{n}}}}}{{{{6.5}^{\rm{n}}}}} = \lim \frac{{{5^n}.\left[ {4 + 4.{{\left( {\frac{2}{5}} \right)}^{\rm{n}}}} \right]}}{{{{6.5}^n}}} = \lim \frac{{4 + 4.{{\left( {\frac{2}{5}} \right)}^{\rm{n}}}}}{6} = \frac{{4 + 4.0}}{6} = \frac{2}{3}\).
g) \(\lim \frac{{2 + \frac{4}{{{n^3}}}}}{{{6^{\rm{n}}}}} = \lim \left( {2 + \frac{4}{{{{\rm{n}}^3}}}} \right).\lim {\left( {\frac{1}{6}} \right)^{\rm{n}}} = \left( {2 + 0} \right).0 = 0\).
tính tổng dãy số:
a, A= 1 . 2 + 2 .3 + 3 . 4 + ... + n . (n+1)
b, B= 1 . 2 . 3 + 2 . 3 . 4 + 3 . 4 . 5 + ... + n . (n+1) . (n+2)
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