Prove that : Find the `4^("th")` term from the end in the expansion `(2a+5b)^(8)`.

Answer

To find the 4th term from the end in the expansion of \((2a + 5b)^8\), we can follow these steps:

### Step 1: Determine the total number of terms
In the expansion of \((x + y)^n\), the total number of terms is given by \(n + 1\). Here, \(n = 8\), so the total number of terms is:
\[ n + 1 = 8 + 1 = 9 \]

**Hint:** Remember that the number of terms in a binomial expansion is always one more than the exponent.

### Step 2: Identify the position of the term from the end
To find the 4th term from the end, we can count backwards from the total number of terms. The 4th term from the end corresponds to the 6th term from the beginning (since \(9 - 4 + 1 = 6\)).

**Hint:** Counting terms from the end can be done by subtracting the desired position from the total number of terms.

### Step 3: Use the binomial theorem to find the 6th term
The general term (or \(r + 1\)th term) in the expansion of \((x + y)^n\) is given by:
\[ T_{r+1} = \binom{n}{r} x^{n-r} y^r \]
For the 6th term, we have \(r = 5\) (since \(r + 1 = 6\)), \(n = 8\), \(x = 2a\), and \(y = 5b\). Thus, we can express the 6th term as:
\[ T_6 = \binom{8}{5} (2a)^{8-5} (5b)^5 \]

**Hint:** Make sure to substitute the correct values for \(n\), \(r\), \(x\), and \(y\) in the formula.

### Step 4: Calculate the components
Now, we calculate each part:
1. Calculate \(\binom{8}{5}\):
\[ \binom{8}{5} = \binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 \]

2. Calculate \((2a)^{3}\):
\[ (2a)^{3} = 2^3 a^3 = 8a^3 \]

3. Calculate \((5b)^{5}\):
\[ (5b)^{5} = 5^5 b^5 = 3125b^5 \]

**Hint:** Break down the calculations into smaller parts to avoid mistakes.

### Step 5: Combine the results
Now we can combine all the components to find the 6th term:
\[ T_6 = \binom{8}{5} (2a)^{3} (5b)^{5} = 56 \cdot 8a^3 \cdot 3125b^5 \]
Calculating this gives:
\[ T_6 = 56 \cdot 8 \cdot 3125 \cdot a^3 \cdot b^5 \]
Calculating \(56 \cdot 8 = 448\) and then \(448 \cdot 3125 = 1400000\):
\[ T_6 = 1400000 a^3 b^5 \]

### Final Answer
Thus, the 4th term from the end in the expansion of \((2a + 5b)^8\) is:
\[ \boxed{1400000 a^3 b^5} \]