Question

If $(x+a)^n = \sum_{k=0}^{n} {n \choose k} x^{n-k} a^k$. Let $a_k = k {10 \choose k}$, $b_k = (10-k) {10 \choose k}$ and $A_k = \begin{bmatrix} a_k & 0 \\ b_k & 0 \end{bmatrix}$. If $A = \sum_{k=0}^{9} A_k = \begin{bmatrix} a & 0 \\ b & 0 \end{bmatrix}$.

Find the sum of digits of trace of the matrix A

Answer 1

6

Answer 2

We are given, for each $k$ (where $0 \le k \le 9$):

$$a_k = k \binom{10}{k},\quad b_k = (10-k) \binom{10}{k}$$

The matrices are:

$$A_k = \begin{pmatrix} a_k & 0 \\ b_k & 0 \end{pmatrix}$$

The matrix $A$ is the sum of these matrices:

$$A = \sum_{k=0}^{9} A_k = \begin{pmatrix} a & 0 \\ b & 0 \end{pmatrix} \quad \text{with} \quad a = \sum_{k=0}^{9} a_k,\quad b = \sum_{k=0}^{9} b_k.$$

The trace of $A$ is the sum of its diagonal entries:

$$\text{trace}(A) = a + b.$$

Notice that for each $k$:

$$a_k + b_k = \left[k + (10-k)\right] \binom{10}{k} = 10 \binom{10}{k}.$$

Thus,

$$a + b = \sum_{k=0}^{9} (a_k + b_k) = \sum_{k=0}^{9} 10 \binom{10}{k} = 10 \sum_{k=0}^{9} \binom{10}{k}.$$

Using the binomial theorem, we know:

$$\sum_{k=0}^{10} \binom{10}{k} = 2^{10} = 1024.$$

Since the sum in our case is only from $k=0$ to $k=9$, we subtract the $k = 10$ term:

$$\sum_{k=0}^{9} \binom{10}{k} = 1024 - \binom{10}{10} = 1024 - 1 = 1023.$$

Thus,

$$\text{trace}(A) = 10 \times 1023 = 10230.$$

Finally, the sum of the digits of $10230$ is:

$$1 + 0 + 2 + 3 + 0 = 6.$$