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2. Show that \left(1+x^2+2x\right)^{2n}=\sum _{k=0}^n\binom{2n}{k}x^{2n-k}\left(x+2\right)^{2n-k}.\: 3. It is known that x^{2n-k}\left(x+2\right)^{2n-k}=\binom{2n-k}{0}2^{2n-k}x^{2n-k}+\binom{2n-k}{1}2^{2n-k-1}x^{2n-k+1}+...+\binom{2n-k}{2n-k}2^{0}x^{4n-2k}. Show that \binom{4n}{2n}=\sum...

b) i. Show that for all positive integers n, x\left[\left(1+x\right)^{n-1}+\left(1+x\right)^{n-2}+...+\left(1+x\right)^{2}+\left(1+x\right)+1\right]=\left(1+x\right)^{n}-1. ii. Hence show that for 1≤k≤n, \binom{n-1}{k-1}+\binom{n-2}{k-1}+\binom{n-3}{k-1}+...+\binom{k-1}{k-1}=\binom{n}{k}...

Question and answer that I wrote = is this the correct answer/working? Thanks!

that is a much faster way - i was drawn to use cases because i got baited by the separate positions for the vowels when actually it doesn't matter for vowel arrangement

How many arrangements of the letters in the word ALGEBRAIC are possible if the vowels must occupy the 2nd, 3rd, 5th and 8th positions? ALGEBRAIC is a 9-letter word Vowels are AAEI, consonants are LGBRC I would solve this using cases: AA in 2rd, 3rd - e and i in 5th, 8th, alongisde 5 consonants...

isn't showing steps behind a paywall for 'pro' pricing plan on wolfram?

Oh I get is now - i did the contradiction proof wrong by assuming the wrong statement :oldmad: thanks tywebb

If\:U_1=\int cot^nxdx,\:show\:that\:U_n=\frac{-1}{n-1}cot^{n-1}x-U_{n-2} I'm going to take a while to do latex so here's a picture of my working so far instead: I don't know how to work the integral on the last step. Maybe some hints? (use spoilers) Thanks so much for the help!

like 5uckerberg's correct answer, the general theory behind trig u-sub integration is: If there are all even powers of trig functions, e.g. \int \sin ^2\left(x\right)\cos ^2\left(x\right)dx, then convert everything into cos double angle form, factorise and work from there If there are one or...

on such graph, impact velocity is 10sqrt21 from formula v^2 = xdot^2+ydot^2. I was just wondering, how do you know if we take the positive (as in this case) or the negative velocity? Thanks so much for your help! [edit:] Wait... maybe the answers supposed to be negative 10sqrt21 from...

It is known that z^{3}+\frac{1}{z^{3}}=w^{3}-3w. from this and the quoted solution above (|w^{2}-3|\ge|w|^{2}-3), prove by contradiction that |z^{3}+\frac{1}{z^{3}}|\le2\ \to|z+\frac{1}{z}|\le2. Assuming that |w^{3}-3w|\le2\ \to|w|\le2 to be true, LHS: |w||w^{2}-3|\le2 =...

bruh wtf (although George is probably cooking something up at ANU for international chem olympiad)

What is this course about that's different to specialising in a clinic after MD? Do you have to take it if you want to become a surgeon? When can you even register to take this course (After Undergrad Med Science? After MD?) [general confusion]

true i've been bulking these holidays as well ill add that to my workout schedule :)

i'm here getting neck pain advice from God himself letsgoooooooooo :shoot:

yeah, thats the thing i mostly do my work on the table tho with physical pieces of paper :( i've used a laptop stand and it works a bit tho

I'm a pretty skinny (decently low body fat BMI 18) guy and I'm getting stomach creases from poor posture - I sit down for most of my day and mainly do lots of work on a laptop or with physical notes. Any tips for better posture and fewer stomach creases?

Evaluate \int _0^{\frac{\pi }{4}}\:sec^4x I know if I(n) = \int \:cos^n\left(x\right)dx then I(n) = \frac{1}{n}sin\left(x\right)cos^{n-1}\left(x\right)+\frac{n-1}{n}I\left(n-2\right) but how would we adapt that to sec^4x? Thanks for the help in advance!

Let z be a complex number. Let w=z+\frac{1}{z}. Show that \left|w^2-3\right|>\left|w\right|^2-3. LHS = \left|z^2+2+\frac{1}{z^2}-3\right| = \left|2cos\left(2\theta \right)-1\right| = 4cos^2x-2-1 = \left(2cos\theta \right)^2-3 RHS = \left|z+\frac{1}{z}\right|^2-3 = \left(2cos\theta \right)^2-3...