Harder 3U question (1 Viewer)

apollo1 · Sep 21, 2011

i) Use binomial theorem to expand: $(1+\frac{1}{n})^{k}$

ii) given $e = \lim_{n \to \infty }(1+\frac{1}{n})^{n}$ prove that e can also be written as:
$1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+.......$

i included the first part just in case it was needed to solve question ii)

michaeljennings · Sep 21, 2011

apollo1 said:
i) Use binomial theorem to expand: $(1+\frac{1}{n})^{k}$

ii) given $e = \lim_{n \to \infty }(1+\frac{1}{n})^{n}$ prove that e can also be written as:
$1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+.......$

i included the first part just in case it was needed to solve question ii)

hahahah wow we did this in class the other day but ive forgotten how to do it lol

xV1P3R · Sep 21, 2011

You just used part i, substitute k = n.

Expand each of the combinations like so:

$nC0 + \frac{nC1}{n} + \frac{nC2}{n^{2}} +...\\=1 + \frac{n!}{(n-1)!1!n} + \frac{n!}{(n-2)!2!n^{2}} + ... \\ = 1 + \frac{1}{1!} + \frac{n(n-1)}{2!n^{2}} + ...$

Then take the limit of the whole thing, and all the n's should cancel out to equal 1.
(Probably safer if you included the fourth term as well. I can't be bothered typing so much)

hup · Sep 21, 2011

michaeljennings said:
hahahah wow we did this in class the other day but ive forgotten how to do it lol

$(1+\frac{1}{n})^n=\sum_{k=0}^{n}\binom{n}{k}\frac{1}{n^k}\\ =\sum_{k=0}^{n}\frac{n(n-1)...(n-k+1)(n-k)!}{k!(n-k!)(n^{k})}\\ =\sum_{k=0}^{n}\frac{1}{k!}\times \frac{n}{n}\times \frac{n-1}{n}\times ...\times \frac{n-k+1}{n}\\ =\sum_{k=0}^{n}\frac{1}{k!}(1-\frac{1}{n})(1-\frac{2}{n})...(1-\frac{1 +k}{n})\\ then\ take\ n \to \infty \\ =\sum_{k=0}^{infinity}\frac{1}{k!}$

then follows

apollo1 · Sep 21, 2011

tnks guys. u guys r amazing. and hup definite state rank 4 u.

largarithmic · Sep 21, 2011

hup said:
$(1+\frac{1}{n})^n=\sum_{k=0}^{n}\binom{n}{k}\frac{1}{n^k}\\ =\sum_{k=0}^{n}\frac{n(n-1)...(n-k+1)(n-k)!}{k!(n-k!)(n^{k})}\\ =\sum_{k=0}^{n}\frac{1}{k!}\times \frac{n}{n}\times \frac{n-1}{n}\times ...\times \frac{n-k+1}{n}\\ =\sum_{k=0}^{n}\frac{1}{k!}(1-\frac{1}{n})(1-\frac{2}{n})...(1-\frac{1 +k}{n})\\ then\ take\ n \to \infty \\ =\sum_{k=0}^{infinity}\frac{1}{k!}$

then follows

This thing has a massive problem with rigour towards the end of the sum. (The reasoning, take n->infinity and then just let the 1+k/n term be zero doesnt work when the k youre dealing with is really big). It's probably enough for HSC though. Also theres a minus sign wrong , should be 1 - (k-1)/n in your second last line.

Harder 3U question (1 Viewer)

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