Photo AI

Given $$ f(x) = (2 + 3x)^3, \, |x| < \frac{2}{3} $$ find the binomial expansion of f(x), in ascending powers of x, up to and including the term in x^3 - Edexcel - A-Level Maths Pure - Question 13 - 2013 - Paper 1

Question icon

Question 13

Given--$$-f(x)-=-(2-+-3x)^3,-\,-|x|-<-\frac{2}{3}-$$--find-the-binomial-expansion-of-f(x),-in-ascending-powers-of-x,-up-to-and-including-the-term-in-x^3-Edexcel-A-Level Maths Pure-Question 13-2013-Paper 1.png

Given $$ f(x) = (2 + 3x)^3, \, |x| < \frac{2}{3} $$ find the binomial expansion of f(x), in ascending powers of x, up to and including the term in x^3. Give each ... show full transcript

Worked Solution & Example Answer:Given $$ f(x) = (2 + 3x)^3, \, |x| < \frac{2}{3} $$ find the binomial expansion of f(x), in ascending powers of x, up to and including the term in x^3 - Edexcel - A-Level Maths Pure - Question 13 - 2013 - Paper 1

Step 1

Find the Binomial Expansion

96%

114 rated

Answer

To find the binomial expansion of the function f(x)=(2+3x)3f(x) = (2 + 3x)^3, we apply the Binomial Theorem, which states:

(a+b)n=k=0n(nk)ankbk(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k

In this case:

  • Let ( a = 2 )
  • Let ( b = 3x )
  • Let ( n = 3 )

Now, we can expand:

(2+3x)3=k=03(3k)(2)3k(3x)k(2 + 3x)^3 = \sum_{k=0}^{3} \binom{3}{k} (2)^{3-k} (3x)^{k}

Calculating each term for ( k = 0 ) to ( 3 ):

  1. For ( k = 0 ): (30)(2)3(3x)0=181=8\binom{3}{0} (2)^3 (3x)^0 = 1 \cdot 8 \cdot 1 = 8

  2. For ( k = 1 ): (31)(2)2(3x)1=343x=36x\binom{3}{1} (2)^2 (3x)^1 = 3 \cdot 4 \cdot 3x = 36x

  3. For ( k = 2 ): (32)(2)1(3x)2=329x2=54x2\binom{3}{2} (2)^1 (3x)^2 = 3 \cdot 2 \cdot 9x^2 = 54x^2

  4. For ( k = 3 ): (33)(2)0(3x)3=1127x3=27x3\binom{3}{3} (2)^0 (3x)^3 = 1 \cdot 1 \cdot 27x^3 = 27x^3

Combining all these terms, we get:

f(x)=8+36x+54x2+27x3f(x) = 8 + 36x + 54x^2 + 27x^3

Thus, the binomial expansion of ( f(x) ) is:

f(x)=8+36x+54x2+27x3f(x) = 8 + 36x + 54x^2 + 27x^3

Join the A-Level students using SimpleStudy...

97% of Students

Report Improved Results

98% of Students

Recommend to friends

100,000+

Students Supported

1 Million+

Questions answered

;