Question

Determine el valor de M.
M=1×3+2×4+3×5+...+20×22

Answer 1

PROPIEDAD DE SMITH.V

sea una progresion de segundo orden:

a₁   a₂    a₃   a₄   a₅  .... aₓ

  n      m    p    q  

      r        r     r

x: numero de terminos

cuyo termino general es :

Tx=a₁.C₀ˣ ⁻ ¹ + n.C₁ ˣ ⁻ ¹+r.C₂ ˣ ⁻ ¹

cuya suma de x terminos es:

∑=a₁.C₁ˣ+n.C₂ˣ+r.C₃ˣ

ten en cuenta que la formula de la combinatoria es:

Cⁿₐ=n!/((n-a)!.a!)

tambien

5! =5.4.3.2.1

1!=1

0!=1

20!=20.19.18.17.......1

un factorial es la multiplicacion de los numeros decrecientes sucesivos del numero

RESOLUCION:

analizamos la sucesion

M=1×3+2×4+3×5+4.6...+20×22

M=3 + 8 + 15 + 24+....+440

      +5   +7    +9

          +2    +2

la cantidad de terminos es: 20

x=20

a₁=3

n=5

r=2

x=20

∑=a₁.C₁ˣ+n.C₂ˣ+r.C₃ˣ

∑=3.C₁²⁰+5.C₂²⁰+2.C₃²⁰

∑=3.[20!/((20-1)!1!)]+5.[20!/((20-2)! .2!)]+2.[20!/((20-3)¡.3!)]

∑=3.[20!/(19!.1!)]+5.[20!/(18! .2!)]+2.[20!/(17!.3!)]

∑=3.20+5. (20.19)/2 +2.(20.19.18)/3.2

∑=60+950+2280

∑=3290

la suma de la serie es 3290

Answer 2

Respuesta:

$\begin{lgathered}\begin{lgathered}\displaystyle 3290 \end{lgathered}\end{lgathered}$

Explicación paso a paso

$\begin{lgathered}\begin{lgathered}\displaystyle \\\sum_{i=1}^{n} i = \frac{n(n+1)}{2} \end{lgathered}\end{lgathered}$

$\begin{lgathered}\begin{lgathered}\displaystyle \\\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6} \end{lgathered}\end{lgathered}$

Resuelve:

$\begin{lgathered}\displaystyle M = 1 \! \times \! 3+ 2 \! \times \! 4 + 3 \! \times \! 5 + \cdots + 20 \! \times \! 22 \end{lgathered}$

$\begin{lgathered}\begin{lgathered}\displaystyle \\\sum_{i=1}^{20} i(i+2)= \sum_{i=1}^{20} i^2 + 2i \end{lgathered}\end{lgathered}$

$\begin{lgathered}\begin{lgathered}\displaystyle \\\sum_{i=1}^{20} i^2 \: \: \: +\: \: \: 2 \sum_{i=1}^{20} i \end{lgathered}\end{lgathered}$

$\begin{lgathered}\begin{lgathered}\displaystyle \frac{20(20+1)(40+1)}{6} \: \: \: +\: \: \: 2 \left ( \frac{20(20+1)}{2} \right ) \end{lgathered}\end{lgathered}$

$\begin{lgathered}\begin{lgathered}\displaystyle \frac{20(21)(41)}{6} \: \: \: +\: \: \: 20(21) \end{lgathered}\end{lgathered}$

$\begin{lgathered}\begin{lgathered}\displaystyle 10(7)(41) \: \: \: +\: \: \: 420 \end{lgathered}\end{lgathered}$

$\begin{lgathered}\begin{lgathered}\displaystyle 2870 \: \: \: +\: \: \: 420 \end{lgathered}\end{lgathered}$

$\begin{lgathered}\begin{lgathered}\displaystyle 3290 \end{lgathered}\end{lgathered}$