Mathematical Induction Problem 2 (IB Math HL)

**alireza1992** · June 23rd 2012, 12:27 AM

Hello.
I have another question about mathematical induction. I have problem dealing with factorial notation in mathematical induction problems. Here are two question which I couldn't solve. I'd be grateful if you could help me to figure out how to solve them.
$Mathematical Induction Problem 2 (IB Math HL)-screen-shot-2012-06-23-10.23.42.png$

**Prove It** · June 23rd 2012, 01:41 AM

Originally Posted by alireza1992

Hello.
I have another question about mathematical induction. I have problem dealing with factorial notation in mathematical induction problems. Here are two question which I couldn't solve. I'd be grateful if you could help me to figure out how to solve them.
$Click image for larger version. Name: Screen shot 2012-06-23 at 10.23.42.png Views: 17 Size: 34.4 KB ID: 24139$

You wish to prove $\displaystyle \begin{align*} 1 \cdot 1! + 2\cdot 2! + 3\cdot 3! + \dots + n \cdot n! = (n + 1)! - 1 \end{align*}$ .

Base step, where $\displaystyle \begin{align*} n = 1 \end{align*}$ :

$\displaystyle \begin{align*} LHS &= 1 \cdot 1! \\ &= 1 \\ \\ RHS &= (1 + 1)! - 1 \\ &= 1 \\ &= LHS \end{align*}$

Inductive step, assume the statement $\displaystyle \begin{align*} 1 \cdot 1! + 2\cdot 2! + 3 \cdot 3! + \dots + n \cdot n! = (n + 1)! - 1 \end{align*}$ holds true for $\displaystyle \begin{align*} n = k \end{align*}$ , and use this to show it holds true for $\displaystyle \begin{align*} n = k + 1 \end{align*}$ , i.e. show that $\displaystyle \begin{align*} 1 \cdot 1! + 2\cdot 2! + 3\cdot 3! + \dots + k\cdot k! + (k + 1)\cdot (k + 1)! = (k + 2)! - 1 \end{align*}$ .

When we let $\displaystyle \begin{align*} n = k + 1 \end{align*}$ we have

$\displaystyle \begin{align*} LHS &= 1 \cdot 1! + 2 \cdot 2! + 3\cdot 3! + \dots + k \cdot k! + (k + 1)\cdot (k + 1)! \\ &= (k + 1)! - 1 + (k + 1)\cdot (k + 1)! \\ &= (k + 1)! + (k + 1)(k + 1)! - 1 \\ &= (k + 1)!( 1 + k + 1 ) - 1 \\ &= (k + 1)!( k + 2) - 1 \\ &= (k + 2)! - 1 \\ &= RHS \end{align*}$

Q.E.D.

**Reckoner** · June 23rd 2012, 01:49 AM

Originally Posted by alireza1992

Here are two question which I couldn't solve. I'd be grateful if you could help me to figure out how to solve them.

For the first one,

Base case: $1\times1! = 1 = (1+1)! - 1.$

Now, assume that the statement is true for $n = k.$ So, we have

$1\times1! + 2\times2! + 3\times3! +\cdots+ k\times k! = (k+1)! - 1$

$\Rightarrow1\times1! + 2\times2! + 3\times3! +\cdots+ k\times k! + (k+1)(k+1)!$

. . . . $= (k+1)! - 1 + (k+1)(k+1)!$

$\Rightarrow1\times1! + 2\times2! + 3\times3! +\cdots+ k\times k! + (k+1)(k+1)!$

. . . . $= (k+1)! + (k+1)(k+1)! - 1$

. . . . $= (k+1)!\left[1 + (k+1)\right] - 1$

. . . . $= (k+1)!(k+2) - 1$

. . . . $= (k+2)! - 1$

**alireza1992** · June 23rd 2012, 08:04 AM

Thanks a lot!

**GGBcn** · July 7th 2012, 10:22 AM

Proof to part b:

Base: Where n=1:
LHS=1/2!
RHS=((1+1)!-1)/(1+1)!=(2!-1)/2!=1/2!=LHS

Now we assume the statement holds true for n=k, and show that it holds true for n=k+1
So we have
1/2!+2/3!+3/4!+4/5!+⋯+k/(k+1)!=((k+1)!-1)/(k+1)!
And want to show that
1/2!+2/3!+3/4!+4/5!+⋯+k/(k+1)!+((k+1))/(k+2)!=((k+2)!-1)/(k+2)!

LHS=1/2!+2/3!+3/4!+4/5!+⋯+k/(k+1)!+((k+1))/(k+2)!

=((k+1)!-1)/(k+1)!+((k+1))/(k+2)!

=((k+2)!(k+1)!-(k+2)!+(k+1)(k+1)!)/(k+1)!(k+2)!

=((k+2)!(k+1)!-(k+2)(k+1)!+(k+1)(k+1)!)/(k+1)!(k+2)!

=((k+2)!-(k+2)+(k+1))/(k+2)!

=((k+2)!-k+k-2+1)/(k+2)!

=((k+2)!-1)/(k+2)!

Q.E.D.

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