MC-vragen voor 17 plussers (500-600) VARIA

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Volledige Inductie
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V o l l e d i g e i n d u c t i e
Mathematical induction (Engels)
Vollständige Induktion (Duits)
Induction Mathématique (Frans)
Indução matemática (Portugees)
Induzione matematica (Italiaans)
Inducción completa (Spaans)

→ telling vanaf 25 juli 2024 ←


$\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+ \cdots + \frac{1}{n.(n+1)} = \frac{n}{n+1}$	³⁵
$\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\cdots +\frac{1}{(2n-\!1)(2n\!+\!1)}=\frac{n}{2n+1}$	²⁰
$\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{11}+\cdots +\frac{1}{(3n-2)(3n+1)}=\frac{n}{3n+1}$	^J1
$\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+\cdots +\frac{1}{(4n-3)(4n+1)}=\frac{n}{4n+1}$	^F5
$\sum_{i=1}^{n}\; \left ( \frac{1}{[1+(i-1)a].[1+i\, a]} \right ) =\!\frac{n}{an\!+\!1}$	^J7
$\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\cdots +\frac{1}{n}=\frac{1}{2}\left ( 1-\frac{1}{3^n} \right )$	⁴⁴
$\frac{1}{2}-\sum_{i=1}^{n}\: \frac{2}{i\, (i+1)(i+2)} = \frac{1}{(n+1)(n+2)}$	⁴⁵
$\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\cdots + \frac{n}{2^n}=2-\frac{n+2}{2^n}$	¹²
$\frac{3}{2}- \sum_{i=1}^{n}\frac{2}{i.(i+2)} =\frac{1}{n+1}+\frac{1}{n+2}$	⁴⁶
$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+ ... +\frac{1}{2n-1}-\frac{1}{2n}=\frac{1}{n+1}+...+\frac{1}{2n}$	⁴⁸
$\frac{1}{2^2-1}+\frac{1}{4^2-1}-\frac{1}{6^2-1}+...+\frac{1}{(2n)^2-1} =\frac{n}{2n+1}$	⁴⁹
$\frac{1}{x}+\frac{1}{x^2}+\cdots+\frac{1}{x^n}=\frac{1}{x-1}-\frac{1}{x^n(x-1)}$	¹³
$\left ( 1\! +\frac{1}{1} \right ).\left ( 1\! +\frac{1}{2} \right ).\left ( 1\! +\frac{1}{3} \right ).\cdots .\left ( 1\! +\frac{1}{n} \right )=n+1$	⁸⁶
$\frac{1^2}{1.3}+\frac{2^2}{3.5}+\cdots +\frac{n^2}{(2n-1)(2n+1)}=\frac{n.(n+1)}{2.(2n+1)}$	⁹⁵
$\sum_{i=1}^{n}\; \sqrt{1+\frac{1}{i^{\, 2}}+\frac{1}{(\, i+1)^2}}\; =\; \frac{n.(n+2)}{n+1}$	⁹⁹
$\frac{1}{1.2.3}+\frac{1}{2.3.4}+ \frac{1}{3.4.5}+\cdots +\frac{1}{n.(n+1).(n+2)}=\frac{n\, (n+3)}{4\, (n+1)\, (n+2)}$	^D6
$x+x^2+x^3+ \cdots +x^n = \frac{x^{\, n+1}-x}{x-1}$	^A0
$\prod_{i=2}^{n}\left ( 1-\frac{1}{i^^{\, 2}} \right ) = \frac{n+1}{2n}$	^A2
(1+ sec 2α)(1+ sec 4α)....(1+ sec 2ⁿα ) = tan (2ⁿα) . cot α	^A7
$\cos\alpha.\cos2\alpha.\cos 4\alpha.\cdots .\cos\: (2^{n}\alpha )=\frac{\sin\: (2^{n+1}\alpha ) }{2^{n+1}.\sin \alpha }$	^A8
t₁ = 1, t_n+1 = 3t_n− 1 ⇒ t_n = (3ⁿ⁻¹ + 1)	^B4
$\frac{1}{a(a\! +\! 1)}+\frac{1}{(a\! +\! 1)(a\! +\! 2)}+\cdots +\frac{1}{(a\! +\!n-\! 1)(a\! +\!n)} = \frac{n}{a\, (a+n)}$	^D2
$\frac{1^2}{1.3}+\frac{2^2}{3.5}+\cdots +\frac{n^2}{(2n-1)(2n+1)}\: =\: \frac{n\, (n+1)}{2(2n+1)}$	^D5
$\sin x + \sin 2x +\cdots +\sin \, nx = \frac{\sin \frac{(n+1)x}{2}\cdot \sin \frac{nx}{2}} {\sin \frac{x}{2}}$	^E3
$t_1\!=2,\, t_{n+1}=\frac{t_n}{t_n+1} \: \Rightarrow \; t_n=\frac{2}{2n-1}$	^E8
$\frac{2}{1.3}+\frac{2}{2.4}+\frac{2}{3.5}+\cdots \frac{2}{n(n\!+\!2)}=\frac{3}{2}-\frac{1}{n+1}-\frac{1}{n+2}$	^I4
$\frac{1}{2.3}+\frac{1}{5.8}+\frac{1}{8.11}+\cdots +\frac{1}{(3n-\!1)(3n+\!2)} =\: \frac{n}{6n+4}$	¹⁸
$\left ( 1-\frac{1}{2^{\, 2}} \right ) \left ( 1-\frac{1}{3^{\, 2}} \right ) \cdots \left ( 1-\frac{1}{n^{\, 2}} \right )\:=\: \frac{n+1}{2n}$	^J6