binomial q (1 Viewer)

[erm]

Hey guys;

find, as a rational number, the coefficient of x in the expansion of (x² + (1/2x))⁸

I know this is an easy question, but i can't seem to do it.

My thinking:

coefficient of x means the coefficient of the 1st term i.e. ⁿC₀ right (but it doesn't work out)?? or am i doing this wrong?

anyone with help would be great.

 

[lolokay]

you want to find 2*n + -1*(8-n) = 1
3n = 9
n = 3
ie you wan the coefficient of the 3rd term

edit: stuffed that. you want the term n from the end - there are 9 terms so you want the 6th term

 

[lyounamu]

Hey guys;

find, as a rational number, the coefficient of x in the expansion of (x² + (1/2x))⁸

I know this is an easy question, but i can't seem to do it.

My thinking:

coefficient of x means the coefficient of the 1st term i.e. ⁿC₀ right (but it doesn't work out)?? or am i doing this wrong?

anyone with help would be great.

T<SUB>r+1</SUB> = 8C<SUB>r</SUB>(x<SUP>2</SUP>)<SUP>8-r</SUP>(1/2x)<SUP>r<?xml:namespace prefix = o ns = "urn:schemas-microsoft-comfficeffice" /><o></o></SUP>
= 8C<SUB>r</SUB>x<SUP>16-2r</SUP>(1/2)<SUP>r</SUP>x<SUP>-r<o></o></SUP>
= 1/2<SUP>r . </SUP>8C<SUB>r</SUB>x<SUP>16-3r</SUP>
The power of x is 1 when 16-3r=1
r = 5
So the 6<SUP>th</SUP> term is when the x has the power of 1.
T<SUB>6</SUB> = 8C<SUB>5</SUB>(x<SUP>2</SUP>)<SUP>3</SUP>(1/2x)<SUP>5<o></o></SUP>
= 7/4x

 

[erm]

T<SUB>r+1</SUB> = 8C<SUB>r</SUB>(x<SUP>2</SUP>)<SUP>8-r</SUP>(1/2x)<SUP>rfficeffice" /><O></O></SUP>
<SUP>is that a formatting error or something?</SUP>

= 8C<SUB>r</SUB>x<SUP>16-2r</SUP>(1/2)<SUP>r</SUP>x<SUP>-r<O></O></SUP>
= 1/2<SUP>r . </SUP>8C<SUB>r</SUB>x<SUP>16-3r</SUP>
The power of x is 1 when 16-3r=1
r = 5
So the 6<SUP>th</SUP> term is when the x has the power of 1.
T<SUB>6</SUB> = 8C<SUB>5</SUB>(x<SUP>2</SUP>)<SUP>3</SUP>(1/2x)<SUP>5<O></O></SUP>
= 7/4x

ok i think i get it now. thanks namu and poster above

 

[lyounamu]

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ok i think i get it now. thanks namu and poster above

Yeah. I wrote first on my word document. But...the computer sorta screwed it up. I tried to fix it but I was bombarded with 1048208124 numbers of random scripts that I could not decode so I just left them as they are, sorry!

I guess I need to study about word document before I do anything, lol

 

[vds700]

Yeah. I wrote first on my word document. But...the computer sorta screwed it up. I tried to fix it but I was bombarded with 1048208124 numbers of random scripts that I could not decode so I just left them as they are, sorry!

I guess I need to study about word document before I do anything, lol

lol namu probably easier to just use sup and sub notation

 

[lyounamu]

lol namu probably easier to just use sup and sub notation

sup? sub notation?

What are they?

Sorry, I am quite stupid with some mathematical terms and computer terms (if that's what you are referring to)

 

[3unitz]

sup? sub notation?

What are they?

Sorry, I am quite stupid with some mathematical terms and computer terms (if that's what you are referring to)

[ sup] sup notation [ /sup] = ^{sup notation}

[ sub] sub notation [ /sub] = _{sub notation}

eg. ⁿC_r

 

[lyounamu]

[ sup] sup notation [ /sup] = ^{sup notation}

[ sub] sub notation [ /sub] = _{sub notation}

eg. ⁿC_r

Ah.... sweet! thanks.

 

[erm]

Got a couple more q's i need help with:

1.
i)Write down the Binomial expansion of (1+x)ⁿ. Hence show that nC0 + nC1 + nC2+...+ nCn = 2ⁿ.

2.
i)Find greatest coefficient in expansion of (1/3 + 2x)¹⁸. -->I don't have any problems working out that r is less than or equal to 15 2/7 - i just don't know how to interpret that to determine which coefficient is the greatest.

ii)Given that x=2/7 in the above expansion, show that there are two consecutive terms which are equal in value and greater than all other terms.

 

[lolokay]

Got a couple more q's i need help with:

1.
i)Write down the Binomial expansion of (1+x)ⁿ. Hence show that nC0 + nC1 + nC2+...+ nCn = 2ⁿ.

(1+x)^n = nCo x^0 + nC1 x^1 +...+ nCn x^n
the total number of terms in the expansion before collecting like terms is 2^n - as you are picking either x or 1 (2 choices each time) n times
Therefore nC0 + nC1 + nC2+...+ nCn = 2ⁿ.

there's probably stuff missing for this to fully count as a proof though

 

[3unitz]

(1+x)^n = nCo x^0 + nC1 x^1 +...+ nCn x^n
the total number of terms in the expansion before collecting like terms is 2^n - as you are picking either x or 1 (2 choices each time) n times
Therefore nC0 + nC1 + nC2+...+ nCn = 2ⁿ.

there's probably stuff missing for this to fully count as a proof though

just sub x = 1 into (1+x)^n = nCo x^0 + nC1 x^1 +...+ nCn x^n

 

[conics2008]

Got a couple more q's i need help with:

1.
i)Write down the Binomial expansion of (1+x)ⁿ. Hence show that nC0 + nC1 + nC2+...+ nCn = 2ⁿ.

2.
i)Find greatest coefficient in expansion of (1/3 + 2x)¹⁸. -->I don't have any problems working out that r is less than or equal to 15 2/7 - i just don't know how to interpret that to determine which coefficient is the greatest.

ii)Given that x=2/7 in the above expansion, show that there are two consecutive terms which are equal in value and greater than all other terms.

i) expand then sub x=1 then you will get 2^n= .....
ii) there are two ways to do this, use the formula which is simple and goes like

n-k+1/k * b/a greater or equal to 1

or the long one ( recommended ) find T(k+1) and divide it by T(k)

abit long working out but it will get you to the formula.

iii) i dont get it. Im going to take a stab at it. see how you found the greatest term. when you find it, sub x= 2/7 see that value, go find another value like it ... ( I think )