2 Problems Involving Sequences & Series

[Vertciel]

Hello there,

I am having some trouble with the following two problems and I would appreciate any help. My work is shown below.

Thank you.

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1. With mathematical induction, prove that $\displaystyle 1 \times 1! + 2 \times 2! + ... + n \times n! = (n + 1)! - 1$.

2. Two different numbers, whose product is 125, are 3 consecutive terms in a geometric sequence. At the same time, they are the first, third, and sixth terms of an arithmetic sequence. Find the numbers.

Work :

1. I showed that n(1) was correct:

$\displaystyle 1 \times 1! = (1 + 1)! - 1 = 1$

Then I showed n + 1:

$\displaystyle (1 \times 1!) + ... + (n \times n!) + [n + 1 \times (n + 1)!]$ = $\displaystyle (n + 2)! - 1$

However, I do not know how to solve the following to prove the statement true:

$\displaystyle (n + 1)! - 1 + [(n + 1) \times (n + 1)!]$

2. The three terms are: a, ar, $\displaystyle ar^2$.

Expressing them as part of the arithmetic sequence:

a = a

ar = a + 2d

$\displaystyle ar^2 = a + 5d$

Since their product is 125:

$\displaystyle a(ar)(ar^2) = 125$

$\displaystyle a^{3}r^{3} = 125$

$\displaystyle r = 5/a$

From here, I tried to substitute this into equations but I seem to have thrown myself into a mess of equations, without seeing how I could determine either a or d.

 

[Denis]

ar = a + 2d : a = 2d / (r - 1)

ar^2 = a + 5d : a = 5d / (r^2 - 1)

2d / (r - 1) = 5d / [(r + 1)(r - 1)] : r = 3/2

OK?

 

[soroban]

Hello, Vertciel!

$\displaystyle \text{1. By mathematical induction, prove that:}$

. . \(\displaystyle 1\!\cdot\!1! + 2\!\cdot\!2! + 3\!\cdot\!3! + \hdots + n\!\cdot\!n! \:= \n + 1)! - 1\)

\(\displaystyle \text{Verify }S(1)\!:\;\;1\!\cdot\!1! \:=\:2!-1\quad\hdots\quad \text{True!}\)


\(\displaystyle \text{Assume }S(k)\!:\;\;1\!\cdot\!1! + 2\!\cdot\!2! + 3!\cdot\!3! + \hdots + k\!\cdot\!k! \;=\;(k+1)!-1\)


$\displaystyle \text{Add }(k+1)\!\cdot\!(k+1)!\text{ to both sides:}$

. . \(\displaystyle \underbrace{1\!\cdot\!1! + 2\!\cdot\!2! + 3\!\cdot\!3! + \hdots + (k+1)\!\cdot\!(k+1)!}_{\text{This is the left side of }S(k+1)} \;=\;(k+1)! + (k+1)(k+1)! - 1\)

$\displaystyle \text{On the right side, factor: }\;(k+1)!\,(1 + k+1)-1 \;=\;(k+1)!\,(k+2)-1 \;=\;(k+2)!-1$


\(\displaystyle \text{We have: }\:1\!\cdot\!1! + 2\!\cdot\!2! + 3\!\cdot\!3! + \hdots + (k+1)\!\cdot\!(k+1)! \;\;=\;\;(k+2)!-1\)


\(\displaystyle \text{And we have proved }S(k\!+\!1) \quad\hdots\quad \text{The inductive proof is complete.}\)

 

[Vertciel]

Dear denis and soroban,

Thank you very much for your replies!