How do I finish this summations problem?

Question

I have posted a picture since I don't know how to make the summation symbols with the lower and upper summations on keyboard, sorry about that..

$$\sum_{a=1}^9\sum_{b=0}^9(101a+10b)$$

The answer is $49,500$, and I am not sure how I am supposed to solve this. The book does have the steps, but they are all alien to me because it doesn't explain how they got from one step to the other.

To be precise I am confused about how they managed to bring $101a + 10b$ into the middle of the two summations, turning it into $\sum_{a=1}^9 (10(101a)+10)\sum_{b=0}^9 b$

Here is what I am referring to,

Answer 1

$$\sum_{a=1}^{9}\sum_{b=0}^{9}\left(101a+10b\right)=\sum_{a=1}^{9}\sum_{b=0}^{9}101a+\sum_{a=1}^{9}\sum_{b=0}^{9}10b$$ $$=101\sum_{a=1}^{9}\sum_{b=0}^{9}a+10\sum_{a=1}^{9}\sum_{b=0}^{9}b=101\times10\left(\sum_{a=1}^9a\right)+10\times9\sum_{b=0}^9b=\left(1010\times45\right)+\left(90\times45\right)=49,500$$

In the second summation I first summed over $a$ which is not present and hence $10b$ is added $9$ times. Hence $10 \times 9$.

Answer 2

The $101a$ doesn't depend on $b$, so you can distribute it out. Then the $10b$ doesn't depend on $a$ so it will distribute as well. Do you know $\sum_{i=0}^n i=\frac 12n(n-1)$? Do you see how this applies?

After your "To be precise" edit: The first term is responding to the fact that it doesn't depend on $b$, so is constant in that sum. The factor $10$ comes from the number of terms. The $10$ comes from the common factor on the $10b$ to get the pattern above.

Answer 3

This kind of problems can't resist solving even though they have been solved by others or answered by others. Here is what I did. I am poor at writing math equations on computer so wrote on paper and attached the photo.