Expand `(2+ x)^(5) - (2- x)^(5)` in ascending powers of x and simplify your result.

Answer

To solve the problem of expanding \((2 + x)^5 - (2 - x)^5\) in ascending powers of \(x\) and simplifying the result, we can follow these steps:

### Step 1: Expand \((2 + x)^5\) using the Binomial Theorem

The Binomial Theorem states that:
\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]
In our case, \(a = 2\), \(b = x\), and \(n = 5\).

So, we expand \((2 + x)^5\):
\[ (2 + x)^5 = \sum_{k=0}^{5} \binom{5}{k} 2^{5-k} x^k \]

Calculating each term:
- For \(k = 0\): \(\binom{5}{0} 2^5 x^0 = 1 \cdot 32 \cdot 1 = 32\)
- For \(k = 1\): \(\binom{5}{1} 2^4 x^1 = 5 \cdot 16 \cdot x = 80x\)
- For \(k = 2\): \(\binom{5}{2} 2^3 x^2 = 10 \cdot 8 \cdot x^2 = 80x^2\)
- For \(k = 3\): \(\binom{5}{3} 2^2 x^3 = 10 \cdot 4 \cdot x^3 = 40x^3\)
- For \(k = 4\): \(\binom{5}{4} 2^1 x^4 = 5 \cdot 2 \cdot x^4 = 10x^4\)
- For \(k = 5\): \(\binom{5}{5} 2^0 x^5 = 1 \cdot 1 \cdot x^5 = x^5\)

Combining these, we have:
\[ (2 + x)^5 = 32 + 80x + 80x^2 + 40x^3 + 10x^4 + x^5 \]

### Step 2: Expand \((2 - x)^5\)

Using the same method, we expand \((2 - x)^5\):
\[ (2 - x)^5 = \sum_{k=0}^{5} \binom{5}{k} 2^{5-k} (-x)^k \]

Calculating each term:
- For \(k = 0\): \(\binom{5}{0} 2^5 (-x)^0 = 32\)
- For \(k = 1\): \(\binom{5}{1} 2^4 (-x)^1 = -80x\)
- For \(k = 2\): \(\binom{5}{2} 2^3 (-x)^2 = 80x^2\)
- For \(k = 3\): \(\binom{5}{3} 2^2 (-x)^3 = -40x^3\)
- For \(k = 4\): \(\binom{5}{4} 2^1 (-x)^4 = 10x^4\)
- For \(k = 5\): \(\binom{5}{5} 2^0 (-x)^5 = -x^5\)

Combining these, we have:
\[ (2 - x)^5 = 32 - 80x + 80x^2 - 40x^3 + 10x^4 - x^5 \]

### Step 3: Subtract the two expansions

Now we need to compute:
\[ (2 + x)^5 - (2 - x)^5 \]
This gives:
\[ (32 + 80x + 80x^2 + 40x^3 + 10x^4 + x^5) - (32 - 80x + 80x^2 - 40x^3 + 10x^4 - x^5) \]

Distributing the negative sign:
\[ = 32 + 80x + 80x^2 + 40x^3 + 10x^4 + x^5 - 32 + 80x - 80x^2 + 40x^3 - 10x^4 + x^5 \]

### Step 4: Combine like terms

Now we combine like terms:
- The constant terms: \(32 - 32 = 0\)
- The \(x\) terms: \(80x + 80x = 160x\)
- The \(x^2\) terms: \(80x^2 - 80x^2 = 0\)
- The \(x^3\) terms: \(40x^3 + 40x^3 = 80x^3\)
- The \(x^4\) terms: \(10x^4 - 10x^4 = 0\)
- The \(x^5\) terms: \(x^5 + x^5 = 2x^5\)

Putting it all together, we have:
\[ (2 + x)^5 - (2 - x)^5 = 160x + 80x^3 + 2x^5 \]

### Final Result
The final result in ascending powers of \(x\) is:
\[ 2x^5 + 80x^3 + 160x \]