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- #1

- Thread starter Carla1985
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- #1

- Feb 13, 2012

- 1,704

The point (i) is justified by the series expansion...

$\displaystyle (1+x)^{n}= \sum_{k=0}^{n} \binom {n}{k}\ x^{k}$ (1)

... where all terms are positive and $\displaystyle \binom{n}{2}\ x^{2}$ is the only second term. The other steps are quite simple...

Kind regards

$\chi$ $\sigma$

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- #3

For example, for (i) you could write:

$\displaystyle (1+x)^n=\sum_{k=0}^n{n \choose k}x^k={n \choose 2}x^2+\left(1+nx+\sum_{k=3}^n{n \choose k}x^k \right)$

For $k\ge0$ it must be true that:

$\displaystyle 1+nx+\sum_{k=3}^n{n \choose k}x^k\ge0$

and so by adding $\displaystyle {n \choose 2}x^2$ to both sides, we obtain:

$\displaystyle (1+x)^n\ge{n \choose 2}x^2=\frac{n(n-1)}{2}x^2$

For (ii) you could begin with:

$\displaystyle \frac{n}{2}\ge1$ for $n\ge2$

Now add $\displaystyle \frac{n}{2}$ to both sides:

$\displaystyle n\ge1+\frac{n}{2}$

Subtract through by 1:

$\displaystyle n-1\ge\frac{n}{2}$

Now, multiply through by $\displaystyle \frac{n}{2}$

$\displaystyle \frac{n(n-1)}{2}\ge\frac{n^2}{4}$

From (i) we have:

$\displaystyle (1+x)^n\ge\frac{n(n-1)}{2}x^2$

And so on the right, replacing $\displaystyle \frac{n(n-1)}{2}$ with $\displaystyle \frac{n^2}{4}$ we obtain:

$\displaystyle (1+x)^n\ge\frac{n^4}{4}x^2$

Multiplying through by $\displaystyle \frac{4}{n^2}$ there results:

$\displaystyle \frac{4}{n^2}(1+x)^n\ge x^2$

$\displaystyle x^2\le\frac{4}{n^2}(1+x)^n$ where $n\ge2,\,x\ge0$

Now for (iii), it is just a matter of applying the result of Problem 1(ii) (which you haven't provided).

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- #5

$\displaystyle \lim_{n\to\infty}\frac{1}{\sqrt{n}}=0$

$\displaystyle \lim_{n\to\infty}\sqrt{\frac{1}{n}}=0$

$\displaystyle \sqrt{\lim_{n\to\infty}\frac{1}{n}}=0$

$\displaystyle \lim_{n\to\infty}\frac{1}{n}=0$

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- #6

- Feb 5, 2012

- 1,621

Hi Carla1985,

Thought you might be interested in seeing >>this<<.

Kind Regards,

Sudharaka.