Photo AI

2. (a) Find the first 3 terms, in ascending powers of $x$, of the binomial expansion of (3+bx)^5 where $b$ is a non-zero constant - Edexcel - A-Level Maths Pure - Question 4 - 2011 - Paper 2

Question icon

Question 4

2.-(a)-Find-the-first-3-terms,-in-ascending-powers-of-$x$,-of-the-binomial-expansion-of--(3+bx)^5-where-$b$-is-a-non-zero-constant-Edexcel-A-Level Maths Pure-Question 4-2011-Paper 2.png

2. (a) Find the first 3 terms, in ascending powers of $x$, of the binomial expansion of (3+bx)^5 where $b$ is a non-zero constant. Give each term in its simplest fo... show full transcript

Worked Solution & Example Answer:2. (a) Find the first 3 terms, in ascending powers of $x$, of the binomial expansion of (3+bx)^5 where $b$ is a non-zero constant - Edexcel - A-Level Maths Pure - Question 4 - 2011 - Paper 2

Step 1

Find the first 3 terms, in ascending powers of $x$, of the binomial expansion of (3+bx)^5

96%

114 rated

Answer

To find the first three terms of the binomial expansion (3+bx)5(3 + bx)^5, we use the Binomial Theorem, which states that:

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

Here, let a=3a = 3, b=bxb = bx, and n=5n = 5.

First Term (k=0k=0):

When k=0k=0:

T0=(50)(3)5(bx)0=12431=243T_0 = {5 \choose 0}(3)^{5}(bx)^{0} = 1 \cdot 243 \cdot 1 = 243

Second Term (k=1k=1):

When k=1k=1:

T1=(51)(3)4(bx)1=581bx=405bxT_1 = {5 \choose 1}(3)^{4}(bx)^{1} = 5 \cdot 81 \cdot bx = 405bx

Third Term (k=2k=2):

When k=2k=2:

T2=(52)(3)3(bx)2=1027(bx)2=270b2x2T_2 = {5 \choose 2}(3)^{3}(bx)^{2} = 10 \cdot 27 \cdot (bx)^{2} = 270b^2x^2

Thus, the first three terms in ascending powers of xx are:

243+405bx+270b2x2243 + 405bx + 270b^2x^2

Step 2

Find the value of $b$ given that the coefficient of $x^2$ is twice the coefficient of $x$

99%

104 rated

Answer

From the expansion, the coefficient of xx is 405b405b and the coefficient of x2x^2 is 270b2270b^2.

We are given that the coefficient of x2x^2 is twice the coefficient of xx:

270b2=2(405b)270b^2 = 2(405b)

Simplifying the equation:

270b2=810b270b^2 = 810b

Rearranging:

270b2810b=0270b^2 - 810b = 0

Factoring out bb:

b(270b810)=0b(270b - 810) = 0

The solutions are: b=0b = 0 or 270b=810270b = 810

Since bb is a non-zero constant:

b=810270=3b = \frac{810}{270} = 3

Therefore, the value of bb is:

b=3b = 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

;