Binomial Expansion Of (m+3)^4 Explained

Hey guys! Today, we're diving deep into the fascinating world of binomial expansions, specifically tackling an expression that might look a bit intimidating at first glance:

${ }_4 C_0 m^4 3^0+{ }_4 C_1 m^3 3^1+{ }_4 C_2 m^2 3^2+{ }_4 C_3 m^1 3^3+ { }_4 C_4 m^0 3^4 ?$

This looks like a mouthful, right? But don't worry, we're going to break it down piece by piece. What we're actually looking at here is the expansion of $(m+3)^4$. Recognizing this pattern is key to solving problems like this efficiently. The general form of a binomial expansion comes from the Binomial Theorem, which states that for any non-negative integer $n$:

(x+y)^n = inom{n}{0} x^n y^0 + inom{n}{1} x^{n-1} y^1 + inom{n}{2} x^{n-2} y^2 + ext{...} + inom{n}{n-1} x^1 y^{n-1} + inom{n}{n} x^0 y^n

Where inom{n}{k} (read as "n choose k") is the binomial coefficient, calculated as rac{n!}{k!(n-k)!}. These coefficients are also found in Pascal's Triangle. In our specific case, we have $n=4$, $x=m$, and $y=3$. So, the given expression is indeed the expansion of $(m+3)^4$.

Now, let's get down to finding the equivalent expression from the options provided. We need to calculate each term in the expansion. Remember, $m^0 = 1$ and $3^0 = 1$. The binomial coefficients for $n=4$ are found in the 5th row of Pascal's Triangle (starting with row 0): 1, 4, 6, 4, 1. Alternatively, we can calculate them:

• ${ }_4 C_0 = \frac{4!}{0!(4-0)!} = \frac{4!}{1 \cdot 4!} = 1$
• ${ }_4 C_1 = \frac{4!}{1!(4-1)!} = \frac{4!}{1 \cdot 3!} = 4$
• ${ }_4 C_2 = \frac{4!}{2!(4-2)!} = \frac{4!}{2! \cdot 2!} = \frac{24}{2 \cdot 2} = 6$
• ${ }_4 C_3 = \frac{4!}{3!(4-3)!} = \frac{4!}{3! \cdot 1!} = \frac{24}{6 \cdot 1} = 4$
• ${ }_4 C_4 = \frac{4!}{4!(4-4)!} = \frac{4!}{4! \cdot 0!} = 1$

Let's plug these coefficients and the powers of $m$ and $3$ back into the expansion:

• Term 1: ${ }_4 C_0 m^4 3^0 = 1 \cdot m^4 \cdot 1 = m^4$
• Term 2: ${ }_4 C_1 m^3 3^1 = 4 \cdot m^3 \cdot 3 = 12m^3$
• Term 3: ${ }_4 C_2 m^2 3^2 = 6 \cdot m^2 \cdot 9 = 54m^2$
• Term 4: ${ }_4 C_3 m^1 3^3 = 4 \cdot m^1 \cdot 27 = 108m$
• Term 5: ${ }_4 C_4 m^0 3^4 = 1 \cdot 1 \cdot 81 = 81$

Adding all these terms together, we get the full expansion: $m^4 + 12m^3 + 54m^2 + 108m + 81$.

Now, let's look at the options provided. We're searching for an expression that matches $m^4 + 12m^3 + 54m^2 + 108m + 81$.

A. $4 x^4+12 x^3+54 x^2+108 x+81$ - This option has coefficients that don't match the powers of $m$. It seems to be mixing up the base variable $m$ with the coefficient of the second term, $4$. Also, the first term $4x^4$ is incorrect.

B. $4 x^4+12 x^3+54 x^2+108 x+324$ - Similar to option A, the first term is incorrect, and the last term (324) does not match our calculated 81.

C. $x^4+12 x^3+54 x^2+108 x+81$ - This option looks promising! Let's re-examine our calculated expansion. We found $m^4 + 12m^3 + 54m^2 + 108m + 81$. If we substitute $x$ for $m$, this option becomes $x^4 + 12x^3 + 54x^2 + 108x + 81$. This perfectly matches our calculated expansion, assuming the variable in the options is meant to be $m$ instead of $x$. It's common in multiple-choice questions for the variable name to be slightly different, but the structure and coefficients should align.

Important Note: It seems there might be a typo in the options provided. Option C uses 'x' as the variable instead of 'm'. Assuming this is just a variable name change and the structure is what matters, Option C is the correct equivalent expression. Let's double-check the calculation to be absolutely sure.

We have $(m+3)^4$. Using the Binomial Theorem:

(m+3)^4 = inom{4}{0} m^4 3^0 + inom{4}{1} m^3 3^1 + inom{4}{2} m^2 3^2 + inom{4}{3} m^1 3^3 + inom{4}{4} m^0 3^4

Let's re-calculate the coefficients and terms:

• inom{4}{0} = 1
• inom{4}{1} = 4
• inom{4}{2} = 6
• inom{4}{3} = 4
• inom{4}{4} = 1

And the powers of 3:

• $3^0 = 1$
• $3^1 = 3$
• $3^2 = 9$
• $3^3 = 27$
• $3^4 = 81$

Now, let's reconstruct the terms:

• Term 1: $1 imes m^4 imes 1 = m^4$

• Term 2: $4 imes m^3 imes 3 = 12m^3$

• Term 3: $6 imes m^2 imes 9 = 54m^2$

• Term 4: $4 imes m^1 imes 27 = 108m$

• Term 5: $1 imes m^0 imes 81 = 81$

So, the expanded form is indeed $m^4 + 12m^3 + 54m^2 + 108m + 81$.

Comparing this to the options, and assuming $x$ in option C is equivalent to $m$:

Option C: $x^4+12 x^3+54 x^2+108 x+81$

This matches our result perfectly. The structure of the coefficients and the powers of the variable and the constant term are all in place. The use of $x$ instead of $m$ is a common way to test if you understand the underlying structure of the expansion rather than just matching symbols.

Why are the other options incorrect?

Let's quickly revisit why A and B are definitely wrong. They start with $4x^4$. In our expansion $(m+3)^4$, the first term is $m^4$. The coefficient inom{4}{0} is 1, and the power of $m$ is $m^4$. For the first term to be $4x^4$, either $n$ would have to be different, or the expression would need to be something like $(ax+b)^n$ where $a=2$ and $n=2$, or if $x$ was not the base variable. However, given the structure of the initial expression, it's clearly a binomial expansion of $(m+3)^4$. Therefore, options A and B, which start with $4x^4$, are immediately ruled out.

Option B also has a last term of $324$. Our last term is inom{4}{4} m^0 3^4 = 1 imes 1 imes 81 = 81. So, 324 is incorrect.

Conclusion:

Based on our detailed calculations using the Binomial Theorem, the expression ${ }_4 C_0 m^4 3^0+{ }_4 C_1 m^3 3^1+{ }_4 C_2 m^2 3^2+{ }_4 C_3 m^1 3^3+ { }_4 C_4 m^0 3^4$ is equivalent to $m^4 + 12m^3 + 54m^2 + 108m + 81$. When compared to the provided options, and assuming the variable $x$ in option C is a substitute for $m$, Option C is the correct answer. This problem is a fantastic way to practice recognizing binomial expansions and applying the Binomial Theorem. Keep practicing, guys, and you'll master these in no time!