Buttons

Let $n$ be the number of buttons.

In the worst-case scenario, you have to press at most $(n-1)$ buttons to make sure of the first button of the sequence.

Now to find the second button of the sequence, you have to press the first correct button(you have already known that) and then press another button. If it's wrong, then again press the first correct button and try for another. This time you have to press $((n-2)\times2)$ buttons at most to make sure of the second button of the sequence.

Thus to be sure of the correct sequence of the buttons, we need to press

$\displaystyle \sum_{i=1}^{24} (25-i)\times i = \sum_{i=1}^{24} 25i - i^2 = 25\times \dfrac{24\times25}{2} - \dfrac{24\times25\times49}{6} = 2600 \enspace buttons$

$As \enspace we \enspace know, \\ \boxed{\displaystyle \sum_{i=1}^{n} i = \dfrac{n(n+1)}{2} \\\displaystyle \sum_{i=1}^{n} i^2 = \dfrac{n(n+1)(2n+1)}{6} }$

After being sure of the sequence, we only have to press those $25$ buttons(sequentially) and the door is open.

Total pressed buttons = $2600+25 = \boxed{2625}$

And remember, this is for the worst cases.

first try to think through brain the condition when there are only 4 buttons left how many trials you need to do .

doing this you will get such series : for last button 1 time , for second last 2 time for third last 4 time and for the fourth last 7 time .

as you can see the series is going as : the trials done for button (n) are (n)+(the no. of trials done for any lock after the n lock -1 )

i.e. if you need to find the trials for the fourth last lock you need to do 4+no. of trials done for the next lock -1

i.e 4 + 4-1 = 7

so you can start the series from the last lock i.e 25th and apply the formula accordingly and add it up.