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\(1^2+2^2+3^2+.......+n^2=1\times\left(2-1\right)+2\times\left(3-1\right)+.......+n\left(\left(n+1\right)-1\right)\)=\(\left(1.2+2.3+3.4+......+n\left(n+1\right)\right)-\left(1+2+3+.....+n\right)\)=\(\frac{n\left(n+1\right)\left(n+2\right)-0.1.2}{3}-\frac{n\left(n+1\right)}{2}=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\)
sử dụng qui nạp:
1² + 2² + 3² + 4² + ...+ n² = \(\frac{n\left(n+1\right)\left(2n+1\right)}{6}\) (*)
(*) đúng khi n= 1
giả sử (*) đúng với n= k, ta có:
1² + 2² + 3² + 4² + ...+ k² = \(\frac{k\left(k+1\right)\left(2k+1\right)}{6}\) (1)
ta cm (*) đúng với n = k +1, thật vậy từ (1) cho ta:
1² + 2² + 3² + 4² + ...+ k² + (k + 1)² = \(\frac{k\left(k+1\right)\left(2k+1\right)}{6}\) + (k + 1)²
= (k+1)\(\left(\frac{k\left(2k+1\right)}{6}+\left(k+1\right)\right)\)= (k + 1)\(\frac{2k^2+k+6k+6}{6}\)
= (k + 1)\(\frac{2k^2+7k+6}{6}\) = (k + 1)\(\frac{2k^2+4k+3k+6}{6}\)
= (k + 1)\(\frac{2k\left(k+2\right)+3\left(k+2\right)}{6}\) = (k + 1)\(\frac{\left(k+2\right)\left(2k+3\right)}{6}\)
vậy (*) đúng với n = k + 1, theo nguyên lý qui nạp (*) đúng với mọi n thuộc N*
a: \(\left(n^2+3n-1\right)\left(n+2\right)-n^3+2\)
\(=n^3+2n^2+3n^2+6n-n-2+n^3+2\)
\(=5n^2+5n=5\left(n^2+n\right)⋮5\)
b: \(\left(6n+1\right)\left(n+5\right)-\left(3n+5\right)\left(2n-1\right)\)
\(=6n^2+30n+n+5-6n^2+3n-10n+5\)
\(=24n+10⋮2\)
d: \(=\left(n+1\right)\left(n^2+2n\right)\)
\(=n\left(n+1\right)\left(n+2\right)⋮6\)
\(A=\sqrt[]{1+2+3+...+\left(n-1\right)+n+...+3+2+1}\)
Ta có :
\(1+2+3+...+\left(n-1\right)=\left(n-1\right)+...+3+2+1=\left[\left(n-1\right)-1\right]+1\left(n-1+1\right):2\)
\(=\dfrac{\left(n-1\right)n}{2}\)
\(\Rightarrow A=\sqrt[]{\dfrac{\left(n-1\right)n}{2}.2+n}\)
\(\Rightarrow A=\sqrt[]{\left(n-1\right)n+n}\)
\(\Rightarrow A=\sqrt[]{n^2-n+n}\)
\(\Rightarrow A=\sqrt[]{n^2}\)
\(\Rightarrow A=n\left(n>0\right)\)
\(\Rightarrow dpcm\)
\(\sqrt{1+2+3+..+\left(n-1\right)+n+\left(n-1\right)+...+3+2+1}\)
\(=\sqrt{2\left[1+2+3+...+\left(n-1\right)+n\right]-n}\)
\(=\sqrt{2.\left(n+1\right).n:2-n}\)
\(=\sqrt{n\left(n+1\right)-n}\)
\(=\sqrt{n^2+n-n}\)
\(=\sqrt{n^2}\)
\(=n\)
Đề có cho n >=0 ko bạn?
\(\sqrt{1+2+3+....+\left(n-1\right)+n+\left(n-1\right)+...+3+2+1}\)
\(=\sqrt{2.\left[1+2+3+...+\left(n-1\right)\right]+n}=\sqrt{2.\frac{\left[\left(n-1\right)+1\right]\left(n-1\right)}{2}+n}\)
\(=\sqrt{\left(n-1+1\right)\left(n-1\right)+n}=\sqrt{n.\left(n-1\right)+n}=\sqrt{n^2-n+n}=n\)
a: f(1)=1
=>\(a\cdot1^2+b\cdot1+1=1\)
=>a+b=0
f(-1)=3
=>\(a\cdot\left(-1\right)^2+b\cdot\left(-1\right)+1=3\)
=>a-b=2
mà a+b=0
nên \(a=\dfrac{2+0}{2}=1;b=2-1=1\)
b: a=1 và b=1 nên \(f\left(x\right)=x^2+x+1\)
\(\Leftrightarrow\dfrac{n}{f\left(n\right)}=\dfrac{n}{n^2+n+1}\)
Gọi d=ƯCLN(n^2+n+1;n)
=>\(\left\{{}\begin{matrix}n^2+n+1⋮d\\n⋮d\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}n^2+n+1⋮d\\n\left(n+1\right)⋮d\end{matrix}\right.\)
=>\(\left(n^2+n+1\right)-n\left(n+1\right)⋮d\)
=>\(1⋮d\)
=>d=1
=>ƯCLN(n^2+n+1;n)=1
=>\(\dfrac{n}{f\left(n\right)}=\dfrac{n}{n^2+n+1}\) là phân số tối giản
Đầu tiên, Tính S1=1+2+3+...+n=\(\frac{n\left(n+1\right)}{2}\)
*/ Tính S2=12+22+32+...+n2
Đặt: S2'=1.2+2.3+3.4+...+n(n+1)
=>3S2'=1.2.3+2.3.3+3.4.3+...+n(n+1).3=1.2.3+2.3.(4-1)+3.4.(5-2)+...+n(n+1)[(n+2)−(n−1)]
Nhân ra và rút gọn ta được: 3S2′=n(n+1)(n+2) => S2'=\(\frac{n\left(n+1\right)\left(n+2\right)}{3}\)
Ta lại có: S2′=1.2+2.3+3.4+...+n(n+1)=(12+22+32+...+n2)+(1+2+3+...+n)=S2+S1=S2+\(\frac{n\left(n+1\right)}{2}\)
=> S2=S2'-\(\frac{n\left(n+1\right)}{2}\)=\(\frac{n\left(n+1\right)\left(n+2\right)}{3}\) -\(\frac{n\left(n+1\right)}{2}\)=\(\frac{n\left(n+1\right)\left(2n+1\right)}{6}\)
S3=
\(\frac{1}{\left(n-1\right)n\left(n+1\right)}=\frac{1}{2}.\frac{2}{\left(n-1\right)n\left(n+1\right)}=\frac{1}{2}.\frac{\left(n+1\right)-\left(n-1\right)}{\left(n-1\right)n\left(n+1\right)}\)
\(=\frac{1}{2}\left[\frac{1}{\left(n-1\right)n}-\frac{1}{n\left(n+1\right)}\right]\)
Ta có đpcm.
oh hay quá nhỉ
đề sai