a few random questions.

**pearlyc** · July 31st 2008, 08:36 AM

a few random questions :

1. Given :

$1 + 2 + 3 + ... + n = \frac{n(n+1)}{2}$

(a) Prove this results from first principles.

2. Show from first principles that,

$\sum^n_{i=1} x^i = x\frac{1-x^n}{1-x}$

Hence give an expression of $\sum^{n-1}_{i=0} x^i$

3. Let

$g(t) = exp$ $\{$ $\alpha +$ $\beta$ $t$ $\}$

Given that g(10) = 8.1882 and g(20) = 60.3403, calculate g(15).

4. Consider a function f such that f(11) = 1.234 and f(12)=2.345. Assuming that the function is linear over the interval from 11 to 12, calculate f(11.36).

Someone would please kindly guide me on how to do the questions! Thanks in advance!

**CaptainBlack** · July 31st 2008, 09:23 AM

Originally Posted by pearlyc

a few random questions :

1. Given :
$1 + 2 + 3 + ... + n = \frac{n(n+1)}{2}$

(a) Prove this results from first principles

This depends on what you mean by first principles. Does induction count?

What about this:

Let:

$S(n)=1 + 2 + 3 + ... + n$

Then:

$(\Delta S)(n)=S(n)-S(n-1)=n,\ \ \ n\ge 2$

$ (\Delta^2 S)(n)=(\Delta S)(n)-(\Delta S)(n-1)=1, \ \ \ n\ge 3 $

That is the second differences of $S(n)$ are constant, which means that $S(n)$ is a quadratic in $n$ .

So let:

$ S(n)=a n^2 + b n+c $

Now:

$S(1)=a+b+c=1$
$S(2)=4a+2b+c=3$
$S(3)=9a+3b+c=6$

Thus we have a set of three simultaneous equations in $a$ , $b$ and $c$ which we solve to find the quadratic for $S(n).$

RonL

**Soroban** · July 31st 2008, 09:37 AM

Hello, pearlyc!

3. Let $g(t) \:= \:e^{\alpha +\beta\,t}$

Given that: . $g(10) = 8.1882\,\text{ and }\,g(20) = 60.3403$

. . calculate $g(15).$

We have: . $\begin{array}{cccccc}g(10)\:=\:8.1882 & \Rightarrow & e^{(\alpha+10\beta)} &=& 8.1882 & {\color{blue}[1]} \\ g(20)\:=\:60.3402 & \Rightarrow & e^{(\alpha+20\beta)} &=& 60.3402 & {\color{blue}[2]}\end{array}$

Divide [2] by [1]: . $\frac{e^{(\alpha + 20\beta)}}{e^{(\alpha+10\beta)}} \;=\;\frac{60.3402}{8.1882} \quad\Rightarrow\quad e^{10\beta} \:=\:7.369165384$

Then: . $10\beta \:=\:\ln(7.369165384)$

. . Hence: . $\beta \:=\:0.199730445 \quad\Rightarrow\quad\boxed{\beta \:\approx\:0.2}$

So far: . $g(t) \;=\;e^{(\alpha+0.2t)}$

Then [1] becomes: . $g(10) \;=\;e^{\alpha +2} \;=\;8.1882 \quad\Rightarrow\quad \alpha + 2 \:=\:\ln(8.1882)$

. . Hence: . $\alpha \:=\:0.102694093\quad\Rightarrow\quad \boxed{\alpha \:\approx\:0.1}$

And we have: . $g(t) \;=\;e^{0.1 +0.2t}$

Therefore: . $g(15)\;=\;e^{0.1 + 3} \:=\:e^{3.1} \:=\:22.19795128 \:\approx\:22.2$

**CaptainBlack** · July 31st 2008, 10:07 AM

Originally Posted by pearlyc

2. Show from first principles that,

$\sum^n_{i=1} x^i = x\frac{1-x^n}{1-x}$

Hence give an expression of $\sum^{n-1}_{i=0} x^i$

$(1-x)\sum^n_{i=1} x^i = \sum_{i=1}^n (x^i-x^{i+1})=(x^1-x^2) +(x^2-x^3) + ... +(x^{n-1}-x^n)+(x^n-x^{n+1})$

Collecting like powers:

$(1-x)\sum^n_{i=1} x^i = x^1-x^{n+1}=x(1-x^n)$

RonL

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