Trial p question (1 Viewer)

[Run hard@thehsc]

How can we solve this easily?

 

[yanujw]

4!, 5!, 6! etc. are all multiples of 4. So each of these terms is equal to 1. There are 2017 of these terms
Also;
i^1 = i
i^2 = -1
i^6 = -1

So the sum is 2015 + i
Answer: B

 

[5uckerberg]

View attachment 35085
How can we solve this easily?

For this problem there is a very clever way of approaching this Q focus on focus on the first 4 terms because you will immediately get I-1-1+1+1 till you get to the 2020 factorial power because every power coming after 4! has a multiple of 4 note that there are 2017 terms because you know 2020-3=2017 and you will have 2017-2+I=2015+i

 

[=)(=]

Ahh i saw this in the baulko trial paper the key thing is knowing that i^4n where n is an integer is always 1