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3. (a) Find the first 4 terms of the binomial expansion, in ascending powers of $x$, of \[ \left(1+\frac{x}{4}\right)^8 \] giving each term in its simplest form - Edexcel - A-Level Maths Pure - Question 4 - 2012 - Paper 4
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3. (a) Find the first 4 terms of the binomial expansion, in ascending powers of $x$, of \[ \left(1+\frac{x}{4}\right)^8 \] giving each term in its simplest form. (... show full transcript
Worked Solution & Example Answer:3. (a) Find the first 4 terms of the binomial expansion, in ascending powers of $x$, of \[ \left(1+\frac{x}{4}\right)^8 \] giving each term in its simplest form - Edexcel - A-Level Maths Pure - Question 4 - 2012 - Paper 4
Step 1
Find the first 4 terms of the binomial expansion
Answer
To find the first four terms of the binomial expansion of (\left(1+\frac{x}{4}\right)^8), we can use the Binomial Theorem, which states:
[ (a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k ]
In this case, let (a = 1), (b = \frac{x}{4}), and (n = 8). Thus, we have:
-
For k = 0:
(\binom{8}{0} (1)^{8} \left(\frac{x}{4}\right)^{0} = 1) -
For k = 1:
(\binom{8}{1} (1)^{7} \left(\frac{x}{4}\right)^{1} = 8 \cdot \frac{x}{4} = 2x) -
For k = 2:
(\binom{8}{2} (1)^{6} \left(\frac{x}{4}\right)^{2} = \frac{8\cdot7}{2} \cdot \left(\frac{x^2}{16}\right) = \frac{56}{16} x^2 = 3.5 x^2) -
For k = 3:
(\binom{8}{3} (1)^{5} \left(\frac{x}{4}\right)^{3} = \frac{8\cdot7\cdot6}{6} \cdot \left(\frac{x^3}{64}\right) = 56 \cdot \frac{x^3}{64} = \frac{56}{64} x^3 = 0.875 x^3)
Thus, the first four terms of the expansion are: [ 1 + 2x + 3.5x^2 + 0.875x^3 ]
Step 2
Use your expansion to estimate the value of (1.025)^8
Answer
To estimate the value of ((1.025)^8), we can represent (1.025) as (1 + 0.025), recognizing that (x = 0.025):
Using the expansion we derived: [ 1 + 2(0.025) + 3.5(0.025)^2 + 0.875(0.025)^3 ]
Calculating each term:
- First term: (1)
- Second term: (2 \cdot 0.025 = 0.05)
- Third term: (3.5 \cdot (0.025)^2 = 3.5 \cdot 0.000625 = 0.0021875)
- Fourth term: (0.875 \cdot (0.025)^3 = 0.875 \cdot 0.000015625 = 0.000013671875)
Adding these values together: [ 1 + 0.05 + 0.0021875 + 0.000013671875 \approx 1.052201171875 ]
Rounding to four decimal places, we find: [ (1.025)^8 \approx 1.0522 ]