# Year 1 AS Maths Edexcel question - binomial expansion

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Please can anyone help me with this question..I'm a bit stuck.

When (1-3/2x)^p is expanded in ascending powers of x, the coefficient of x is -24.

a)find the value of p

Many thanks, much appreciated

When (1-3/2x)^p is expanded in ascending powers of x, the coefficient of x is -24.

a)find the value of p

Many thanks, much appreciated

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#2

(Original post by

Please can anyone help me with this question..I'm a bit stuck.

When (1-3/2x)^p is expanded in ascending powers of x, the coefficient of x is -24.

a)find the value of p

Many thanks, much appreciated

**WingsDan**)Please can anyone help me with this question..I'm a bit stuck.

When (1-3/2x)^p is expanded in ascending powers of x, the coefficient of x is -24.

a)find the value of p

Many thanks, much appreciated

(1 + x)^n = 1 + nx/1! + [n(n - 1)x^2]/2! + [n(n - 1)(n - 2)x^3]/3! + ...

Replace x with (-3/2)x, n = p

Alternatively, using the binomial theorem

(a + b)^n = a^n + (nC1)a^(n-1)b + (nC2)a^(n-2)b^2 + … + (nCn-1)ab^(n-1) + b^n

Where (nCr) = n!/(r! (n - r)!), 1 = a, b = (-3x/2)

See: https://revisionmaths.com/advanced-l...inomial-series

You should be able to figure out the rest.

Last edited by MindMax2000; 6 days ago

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(Original post by

Using the binomial theorem for (1 + x)^n:

(1 + x)^n = 1 + nx/1! + [n(n - 1)x^2]/2! + [n(n - 1)(n - 2)x^3]/3! + ...

Replace x with (-3/2)x, n = p

Alternatively, using the binomial theorem

(a + b)^n = a^n + (nC1)a^(n-1)b + (nC2)a^(n-2)b^2 + … + (nCn-1)ab^(n-1) + b^n

Where (nCr) = n!/(r! (n - r)!), 1 = a, b = (-3x/2)

See: https://revisionmaths.com/advanced-l...inomial-series

You should be able to figure out the rest.

**MindMax2000**)Using the binomial theorem for (1 + x)^n:

(1 + x)^n = 1 + nx/1! + [n(n - 1)x^2]/2! + [n(n - 1)(n - 2)x^3]/3! + ...

Replace x with (-3/2)x, n = p

Alternatively, using the binomial theorem

(a + b)^n = a^n + (nC1)a^(n-1)b + (nC2)a^(n-2)b^2 + … + (nCn-1)ab^(n-1) + b^n

Where (nCr) = n!/(r! (n - r)!), 1 = a, b = (-3x/2)

See: https://revisionmaths.com/advanced-l...inomial-series

You should be able to figure out the rest.

a

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#4

**WingsDan**)

Please can anyone help me with this question..I'm a bit stuck.

When (1-3/2x)^p is expanded in ascending powers of x, the coefficient of x is -24.

a)find the value of p

Many thanks, much appreciated

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#5

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yes, so I got 1^p+(pC1)1^p-1(-3/2x)+(pC2)^p-2(-3/2x)^2...but im not sure how to expand.. how would I know what "r" to do it up to? (ie, in this case I done it up to 2)

**WingsDan**)yes, so I got 1^p+(pC1)1^p-1(-3/2x)+(pC2)^p-2(-3/2x)^2...but im not sure how to expand.. how would I know what "r" to do it up to? (ie, in this case I done it up to 2)

btw:

p!/(r!(n-r!) for (pC1) is p!/(1(p-1!). That means:

[p(p-1)(p-2)(p-3)...(3)(2)(1)]/[1(p-1)(p-2)(p-3)....(3)(2)(1)]. If you cancel out the common factors from the numerators and denominators, it will make life a lot easier for yourself.

Also 1 to the power of anything doesn't change the answer

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(Original post by

is it 1 - 3/(2x) or 1 - 1.5x in the bracket ?

**the bear**)is it 1 - 3/(2x) or 1 - 1.5x in the bracket ?

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#7

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(1-(3/2x)^p

**WingsDan**)(1-(3/2x)^p

(Original post by

Looking at your question (quoted above), the coefficient of x is -24. Do you need to know what r is or do it all the way up to r?

btw:

p!/(r!(n-r!) for (pC1) is p!/(1(p-1!). That means:

[p(p-1)(p-2)(p-3)...(3)(2)(1)]/[1(p-1)(p-2)(p-3)....(3)(2)(1)]. If you cancel out the common factors from the numerators and denominators, it will make life a lot easier for yourself.

Also 1 to the power of anything doesn't change the answer

**MindMax2000**)Looking at your question (quoted above), the coefficient of x is -24. Do you need to know what r is or do it all the way up to r?

btw:

p!/(r!(n-r!) for (pC1) is p!/(1(p-1!). That means:

[p(p-1)(p-2)(p-3)...(3)(2)(1)]/[1(p-1)(p-2)(p-3)....(3)(2)(1)]. If you cancel out the common factors from the numerators and denominators, it will make life a lot easier for yourself.

Also 1 to the power of anything doesn't change the answer

**MindMax2000**)

Using the binomial theorem for (1 + x)^n:

(1 + x)^n = 1 + nx/1! + [n(n - 1)x^2]/2! + [n(n - 1)(n - 2)x^3]/3! + ...

Replace x with (-3/2)x, n = p

Alternatively, using the binomial theorem

(a + b)^n = a^n + (nC1)a^(n-1)b + (nC2)a^(n-2)b^2 + … + (nCn-1)ab^(n-1) + b^n

Where (nCr) = n!/(r! (n - r)!), 1 = a, b = (-3x/2)

See: https://revisionmaths.com/advanced-l...inomial-series

You should be able to figure out the rest.

(pC1)1^(p-1)(-3x/2) = -24x

Where (pC1) = p!/(1(p-1!) = [p(p-1)(p-2)(p-3)...(3)(2)(1)]/[1(p-1)(p-2)(p-3)....(3)(2)(1)] = p

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