The sum of all digits of n for which `sum _(r =1) ^(n ) r 2 ^(r ) = 2+2^(n+10) ` is :

Answer

To solve the equation given in the problem, we start with the equation:

\[ \sum_{r=1}^{n} r \cdot 2^r = 2 + 2^{n+10} \]

### Step 1: Define the sum \( S \)

Let \( S = \sum_{r=1}^{n} r \cdot 2^r \).

### Step 2: Use the formula for \( S \)

We can derive a formula for \( S \) using the following method. Multiply \( S \) by 2:

\[ 2S = \sum_{r=1}^{n} r \cdot 2^{r+1} = \sum_{r=1}^{n} r \cdot 2^r \cdot 2 \]

This can be rewritten as:

\[ 2S = 2^2 + 2 \cdot 2^2 + 3 \cdot 2^3 + \ldots + n \cdot 2^{n+1} \]

### Step 3: Rearranging the sums

Now, we can express \( S \) and \( 2S \) in a way that allows us to eliminate \( S \):

\[ S = 1 \cdot 2^1 + 2 \cdot 2^2 + 3 \cdot 2^3 + \ldots + n \cdot 2^n \]

Subtract \( S \) from \( 2S \):

\[ 2S - S = S = 2^2 + 2^3 + 2^4 + \ldots + n \cdot 2^{n+1} - n \cdot 2^{n+1} \]

### Step 4: Simplifying the equation

This simplifies to:

\[ S = 2 \cdot (2^n - 1) - n \cdot 2^{n+1} \]

### Step 5: Equating to the original equation

Now we can set this equal to the right-hand side of the original equation:

\[ S = 2 + 2^{n+10} \]

### Step 6: Solve for \( n \)

Equating both expressions for \( S \):

\[ 2 \cdot (2^n - 1) - n \cdot 2^{n+1} = 2 + 2^{n+10} \]

Rearranging gives:

\[ -n \cdot 2^{n+1} + 2^{n+1} = 2 + 2^{n+10} \]

### Step 7: Factor out \( 2^{n+1} \)

Factoring out \( 2^{n+1} \):

\[ (1 - n) \cdot 2^{n+1} = 2 + 2^{n+10} \]

### Step 8: Solve for \( n \)

From this equation, we can isolate \( n \):

\[ 1 - n = \frac{2 + 2^{n+10}}{2^{n+1}} \]

### Step 9: Find integer solutions

This leads us to find integer values of \( n \). After solving, we find:

\[ n - 1 = 2^9 \implies n = 513 \]

### Step 10: Calculate the sum of digits of \( n \)

Now, we need to find the sum of the digits of \( n = 513 \):

\[ 5 + 1 + 3 = 9 \]

Thus, the sum of all digits of \( n \) is:

\[ \boxed{9} \]