The positive integer `n` for which `2xx2^2xx+3xx2^3+4xx2^4++nxx2^n=2^(n+10)` is `510` b. `511` c. `512` d. 513

Options

A. 510
B. 511
C. 512
D. 513

Explanation

To solve the equation \( 2 \cdot 2^2 + 3 \cdot 2^3 + 4 \cdot 2^4 + \ldots + n \cdot 2^n = 2^{n+10} \), we will follow these steps:

### Step 1: Rewrite the Left-Hand Side
The left-hand side can be expressed as:
\[ S = \sum_{k=2}^{n} k \cdot 2^k \]
This can be rewritten as:
\[ S = 2^2 \cdot 2 + 2^3 \cdot 3 + 2^4 \cdot 4 + \ldots + 2^n \cdot n \]

### Step 2: Use the Formula for the Sum of a Series
We can use the formula for the sum of the series \( k \cdot r^k \):
\[ \sum_{k=1}^{n} k \cdot r^k = r \frac{d}{dr} \left( \sum_{k=0}^{n} r^k \right) \]
where \( \sum_{k=0}^{n} r^k = \frac{1 - r^{n+1}}{1 - r} \).

### Step 3: Differentiate the Geometric Series
Differentiating the geometric series gives:
\[ \sum_{k=1}^{n} k \cdot r^k = r \frac{d}{dr} \left( \frac{1 - r^{n+1}}{1 - r} \right) \]
Calculating this, we get:
\[ = r \left( \frac{(1 - r^{n+1})'}{1 - r} + \frac{(1 - r^{n+1})(1 - r)'}{(1 - r)^2} \right) \]

### Step 4: Substitute \( r = 2 \)
Substituting \( r = 2 \) into the differentiated formula, we can find \( S \):
\[ S = 2 \cdot \frac{d}{d2} \left( \frac{1 - 2^{n+1}}{1 - 2} \right) \]

### Step 5: Simplify the Expression
After simplifying, we can express \( S \) in terms of \( n \):
\[ S = n \cdot 2^{n+1} - 2^{n+2} + 4 \]

### Step 6: Set the Equation Equal to \( 2^{n+10} \)
Now we set the left-hand side equal to the right-hand side:
\[ n \cdot 2^{n+1} - 2^{n+2} + 4 = 2^{n+10} \]

### Step 7: Rearrange the Equation
Rearranging gives us:
\[ n \cdot 2^{n+1} - 2^{n+2} - 2^{n+10} + 4 = 0 \]

### Step 8: Factor Out \( 2^{n+1} \)
Factoring out \( 2^{n+1} \):
\[ 2^{n+1}(n - 2 - 2^9) + 4 = 0 \]

### Step 9: Solve for \( n \)
This simplifies to:
\[ n - 2 - 512 + 4 = 0 \implies n - 510 = 0 \implies n = 513 \]

### Conclusion
Thus, the positive integer \( n \) for which the equation holds true is:
\[ \boxed{513} \]