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Find the first 4 terms, in ascending powers of $x$, of the binomial expansion of $$ \left( 2 - \frac{1}{2} x \right) ^8 $$ giving each term in its simplest form. - Edexcel - A-Level Maths Pure - Question 5 - 2013 - Paper 5

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Find-the-first-4-terms,-in-ascending-powers-of-$x$,-of-the-binomial-expansion-of---$$-\left(-2---\frac{1}{2}-x-\right)-^8-$$-giving-each-term-in-its-simplest-form.-Edexcel-A-Level Maths Pure-Question 5-2013-Paper 5.png

Find the first 4 terms, in ascending powers of $x$, of the binomial expansion of $$ \left( 2 - \frac{1}{2} x \right) ^8 $$ giving each term in its simplest form.

Worked Solution & Example Answer:Find the first 4 terms, in ascending powers of $x$, of the binomial expansion of $$ \left( 2 - \frac{1}{2} x \right) ^8 $$ giving each term in its simplest form. - Edexcel - A-Level Maths Pure - Question 5 - 2013 - Paper 5

Step 1

Find the first term using the Binomial Theorem

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Answer

The first term of the expansion is given by the formula:

(n0)anb0=(80)(2)8(12x)0=256\binom{n}{0} a^n b^0 = \binom{8}{0} \left(2\right)^8 \left(-\frac{1}{2}x\right)^0 = 256

Thus, the first term is 256.

Step 2

Find the second term

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Answer

The second term is:

(81)an1b1=(81)(2)7(12x)1=8128(12x)=512x\binom{8}{1} a^{n-1} b^1 = \binom{8}{1} \left(2\right)^7 \left(-\frac{1}{2}x\right)^1 = 8 \cdot 128 \cdot \left(-\frac{1}{2}x\right) = -512x

So, the second term is -512x.

Step 3

Find the third term

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Answer

The third term can be calculated as:

(82)an2b2=(82)(2)6(12x)2=286414x2=448x2\binom{8}{2} a^{n-2} b^2 = \binom{8}{2} \left(2\right)^6 \left(-\frac{1}{2}x\right)^2 = 28 \cdot 64 \cdot \frac{1}{4} x^2 = 448x^2

Thus, the third term is 448x².

Step 4

Find the fourth term

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Answer

The fourth term is:

(83)an3b3=(83)(2)5(12x)3=563218x3=224x3\binom{8}{3} a^{n-3} b^3 = \binom{8}{3} \left(2\right)^5 \left(-\frac{1}{2}x\right)^3 = 56 \cdot 32 \cdot -\frac{1}{8} x^3 = -224x^3

So, the fourth term is -224x³.

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