Binomial Help PLS (1 Viewer)

pencilspanker

Member
Hey guys having heaps of trouble with binomial, would be great if someone could help with this question!
https://gyazo.com/e0b5251a4a6f2123f58967e4d7049c04
Thanks.
FYI I think the result in ii) you will get n(n-1)2^n-2 i think that is correct atleast thats what i got maybe the question is written wrong
But anyways i need help with iii) just dont know how to start

InteGrand

Well-Known Member
Hey guys having heaps of trouble with binomial, would be great if someone could help with this question!
https://gyazo.com/e0b5251a4a6f2123f58967e4d7049c04
Thanks.
FYI I think the result in ii) you will get n(n-1)2^n-2 i think that is correct atleast thats what i got maybe the question is written wrong
But anyways i need help with iii) just dont know how to start
$\bg_white \noindent By the way, the answer to ii) can't be n(n-1)\cdot 2^{n-2}. If this was the case, then substituting n=1 into your answer gives 0, so the series would have to equal 0 when n=1. But clearly when n=1, the series equals \binom{1}{1}1^{2} = 1 \color{red}\neq\color{black} 0.$

pencilspanker

Member
$\bg_white \noindent By the way, the answer to ii) can't be n(n-1)\cdot 2^{n-2}. If this was the case, then substituting n=1 into your answer gives 0, so the series would have to equal 0 when n=1. But clearly when n=1, the series equals \binom{1}{1}1^{2} = 1 \color{red}\neq\color{black} 0.$
welp gg fuk got ii) wrong as welll yayy

ichila101

for part iii) to start off you are proving mathematical induction so you can start with proving the left side = right side by substituting n = 4 to both sides then go on from there.

a hint for part ii) is to look carefully at the expression and remember you are differentiating with respect to 'x' for the expression n*x*(1+x)^(n-1)

integral95

Well-Known Member
for part iii) to start off you are proving mathematical induction so you can start with proving the left side = right side by substituting n = 4 to both sides then go on from there.

a hint for part ii) is to look carefully at the expression and remember you are differentiating with respect to 'x' for the expression n*x*(1+x)^(n-1)
You don't use mathematical induction in HSC unless otherwise stated.

in iii), you differentiate part i) again, then multiply both sides by x, sub x = -1, then add this with part ii) then divide by 2.