Understanding Binomial Expansion and Coefficients

The question asks us to find the coefficient of the term containing 'y' in the expansion of the expression (2y – 5)³. This involves using the binomial theorem to expand the given expression.

Applying the Binomial Theorem

The binomial theorem provides a formula for expanding expressions of the form (a + b)ⁿ. The general formula is:

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

where $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ is the binomial coefficient.

In our expression, (2y – 5)³, we can identify:

• $a = 2y$
• $b = -5$
• $n = 3$

We need to find the term in the expansion where the power of 'y' is 1 (since we are looking for the coefficient of 'y'). In the general term $\binom{n}{k} a^{n-k} b^k$, the power of 'a' is $(n-k)$. Since $a = 2y$, the power of 'y' is also $(n-k)$. We want $(n-k) = 1$, and we know $n=3$, so $3-k = 1$, which means $k=2$.

Let's calculate the terms for $k=0, 1, 2, 3$ to see the full expansion and confirm the 'y' term.

Step-by-Step Expansion of (2y – 5)³

• For k = 0:

Term $= \binom{3}{0} (2y)^{3-0} (-5)^0 = \binom{3}{0} (2y)^3 (-5)^0$

$\qquad = 1 \cdot (8y^3) \cdot 1 = 8y^3$

• For k = 1:

Term $= \binom{3}{1} (2y)^{3-1} (-5)^1 = \binom{3}{1} (2y)^2 (-5)^1$

$\qquad = 3 \cdot (4y^2) \cdot (-5) = -60y^2$

• For k = 2:

Term $= \binom{3}{2} (2y)^{3-2} (-5)^2 = \binom{3}{2} (2y)^1 (-5)^2$

$\qquad = 3 \cdot (2y) \cdot (25) = 150y$

• For k = 3:

Term $= \binom{3}{3} (2y)^{3-3} (-5)^3 = \binom{3}{3} (2y)^0 (-5)^3$

$\qquad = 1 \cdot (1) \cdot (-125) = -125$

So, the full expansion of (2y – 5)³ is $8y^3 - 60y^2 + 150y - 125$.

Identifying the Coefficient of y

We are looking for the coefficient of the term containing 'y'. From the expansion, the term with 'y' is $150y$. The coefficient of 'y' in this term is 150.

Therefore, the coefficient of y in the expansion of (2y – 5)³ is 150.

The term with 'y' is $150y$, so the coefficient is 150.

Revision Table: Binomial Expansion Concepts

Additional Information: Pascal's Triangle

For small powers like n=3, the binomial coefficients $\binom{n}{k}$ can also be found using Pascal's Triangle. The rows of Pascal's Triangle give the binomial coefficients for increasing powers of n.

• Row 0 (n=0): 1
• Row 1 (n=1): 1, 1
• Row 2 (n=2): 1, 2, 1
• Row 3 (n=3): 1, 3, 3, 1

For (a+b)³, the coefficients are 1, 3, 3, 1. So the expansion is:

$\qquad 1 \cdot a^3 b^0 + 3 \cdot a^2 b^1 + 3 \cdot a^1 b^2 + 1 \cdot a^0 b^3$

Substituting $a=2y$ and $b=-5$:

$\qquad 1 \cdot (2y)^3 (-5)^0 + 3 \cdot (2y)^2 (-5)^1 + 3 \cdot (2y)^1 (-5)^2 + 1 \cdot (2y)^0 (-5)^3$

$\qquad = 1 \cdot 8y^3 \cdot 1 + 3 \cdot 4y^2 \cdot (-5) + 3 \cdot 2y \cdot 25 + 1 \cdot 1 \cdot (-125)$

$\qquad = 8y^3 - 60y^2 + 150y - 125$

This confirms the term $150y$, and its coefficient is 150.