CMR : \(1^2+2^2+3^2+...+n^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\)\(\)
CMR \(\forall n\in\)N* ta có
\(\left(1-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{5}-\frac{1}{6}\right)+...+\left(\frac{1}{2n-1}-\frac{1}{2n}\right)=\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}\)
CMR \(\frac{1.3.5.7............\left(2n-1\right)}{\left(n+1\right).\left(n+2\right).\left(n+3\right)............2n}\)=\(\frac{1}{2^n}\)
CMR : \(\frac{1.3.5.7..............\left(2n-1\right)}{\left(n+1\right)\left(n+2\right)\left(n+3\right)...............2n}\) =\(\frac{1}{^{2^n}}\)
Ta có: \(\frac{1.3.5.7.....\left(2n-1\right)}{\left(n+1\right)\left(n+2\right)\left(n+3\right).....2n}\)
\(=\frac{1.2.3.4..5.6...\left(2n-1\right).2n}{\left(2.4.6....2n\right)\left(n+1\right)\left(n+2\right)\left(n+3\right)....2n}\)
\(=\frac{1.2.3.4.5.6...\left(2n-1\right)}{2^n.1.2.3....n\left(n+1\right)\left(n+2\right)\left(n+3\right)....2n}\)
\(=\frac{1}{2^n}\left(đpcm\right)\)
CMR
\(1\times3+2\times4+3\times5+\left(n-1\right)\left(n+1\right)=\frac{\left(n-1\right)n\left(2n+1\right)}{6}\)
CMR:
\(1^2+2^2+3^2+...+n^2=\frac{n.\left(n+1\right).\left(2n+1\right)}{6}\)
Xét trường hợp n chẵn
12 + 22 + 32 + ... + n2
= [ 12 + 32 + ... + ( n - 1 ) 2 ] + ( 22 + 42 + 62 + ... + n2 )
= \(\frac{\left(n-1\right).n.\left(n+1\right)+n.\left(n+1\right).\left(n+2\right)}{6}\)
= \(\frac{n.\left(n+1\right).\left[\left(n-1\right).\left(n+2\right)\right]}{6}\)
= \(\frac{n.\left(n+1\right).\left(2n+1\right)}{6}\)
Xét trường hợp n lẻ ta có :
12 + 22 + 32 + ... + n2
= ( 12 + 32 + ... + n2 ) + [ 22 + 42 +... + ( n - 1 ) 2 ]
= \(\frac{n.\left(n+1\right).\left(n+2\right)+\left(n-1\right).n.\left(n+1\right)}{6}\)
\(=\frac{n.\left(n+1\right).\left[\left(n+2\right)+\left(n-1\right)\right]}{6}\)
= \(\frac{n.\left(n+1\right).\left(2n+1\right)}{6}\)
Do Not Ask Why
MÌnh không có thời gian trình bày nên bạn thông cảm nha :
Câu hỏi của Đinh Tuấn Việt - Toán lớp 6 - Học toán với OnlineMath
Đặt A = 12 + 22 + 32 +...+ n2
= 1(2 - 1) + 2(3 - 1) + 3(4 - 1) +...+ n[(n + 1) - 1]
= 1.2 - 1.1 + 2.3 - 2.1 + 3.4 - 3.1 +...+ n(n + 1) - n
= [1.2 + 2.3 + 3.4 +...+ n(n + 1)] - (1 + 2 + 3 +... + n)
Đặt B = 1.2 + 2.3 +...+ n(n + 1)
3B = 1.2(3 - 0) + 2.3(4 - 1) +...+ n(n + 1)[(n + 2) - (n - 1)]
3B = 1.2.3 - 0.1.2 + 2.3.4 - 1.2.3 +...+ n(n + 1)(n + 2) - (n - 1)n(n + 1)
3B = n(n + 1)(n + 2)
B = \(\frac{n\left(n+1\right)\left(n+2\right)}{3}\)
Đặt C = 1 + 2 +...+ n
Số số hạng: (n + 1) : 1 + 1 = n (số)
Tổng: \(\frac{\left(n+1\right)n}{2}\)
Thay B và C vào A ta có:
\(A=\frac{n\left(n+1\right)\left(n+2\right)}{3}-\frac{n\left(n+1\right)}{2}=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\)
Bài 1: CMR
a) A = \(\frac{\left(n+1\right).\left(n+2\right)....\left(2n-1\right).\left(2n\right)}{2^n}\) là số nguyên.
b) B = \(\frac{3.\left(n+1\right).\left(n +2\right)...\left(3n-1\right).3n}{3^n}\)là số nguyên.
1. Chứng minh : B = \(\left(1-\frac{2}{6}\right).\left(1-\frac{2}{12}\right).\left(1-\frac{2}{20}\right)...\left(1-\frac{2}{n\left(n+1\right)}\right)>\frac{1}{3}\)
2. cho M = \(\frac{1}{1.\left(2n-1\right)}+\frac{1}{3.\left(2n-3\right)}+\frac{1}{5.\left(2n-5\right)}+...+\frac{1}{\left(2n-3\right).3}+\frac{1}{\left(2n-1\right).1}\)
N = \(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2n-1}\)
Rút gọn \(\frac{M}{N}\)
CMR : A = \(\frac{\left(n+1\right)\left(n+2\right)\left(n+3\right)....\left(2n-1\right).2n}{2^n}\) là một số nguyên
1. Tìm x;y nguyên tố biết : 59x + 46y=2004
2. CMR: \(\frac{1.3.5.7.....\left(2n-1\right)}{\left(n+1\right)\left(n+2\right)\left(n+3\right).....2n}=\frac{1}{2^n}\) với n thuộc N*
a, 59x + 46y = 2004
Vì 2004 là số chẵn, 46y là số chẵn => 59x là số chẵn
=> x là số chẵn, mà x là số nguyên tố
=> x = 2
=> 2.59 + 46y = 2004
=> 46y = 2004 ‐ 118
=> 46y = 1886
=> y = 1886:46 => y = 41
Vậy x = 2; y = 41
Chứng minh rằng với \(n\in N\)* thì:
a, \(1^2+2^2+3^2+...+n^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\)
b, \(1^3+2^3+3^3+...+n^3=\left(\frac{n\left(n+1\right)}{2}\right)^2\)
c, \(n+2\left(n-1\right)+3\left(n-2\right)+...+n=\frac{n\left(n+1\right)\left(n+2\right)}{6}\)