The formula 1 + 2 + 3 + 4 +. + 100 represents the sum of 100 natural numbers starting from 1. Because the formula is relatively long, the writing is not square For the sake of simplicity, we will express it as ∑ (100 above, n = 1 below, N in the middle). Here is the summation sign. Through reading the above materials, we can calculate ∑ (2014 above, n = 1 below, 1 / N of n (n + 1) at the opening in the middle)

The formula 1 + 2 + 3 + 4 +. + 100 represents the sum of 100 natural numbers starting from 1. Because the formula is relatively long, the writing is not square For the sake of simplicity, we will express it as ∑ (100 above, n = 1 below, N in the middle). Here is the summation sign. Through reading the above materials, we can calculate ∑ (2014 above, n = 1 below, 1 / N of n (n + 1) at the opening in the middle)


The final result is 1-1 / (n + 1)!



Observe the following equations: 3 ^ 1 = 3, 3 ^ 2 = 9, 3 ^ 3 = 27, 3 ^ 4 = 81, 3 ^ 5 = 243, 3 ^ 6 = 729, 3 ^ 7 = 2187 You can find 3 ^ n (n is an unnatural number)
Do you know the law of the last number? Write the last number of 3 ^ 2013 according to the law you found


3^1=3,
3^2=9,
3^3=27,
3^4=81,
3^5=243,
3^6=729,
3^7=2187
...
The last position is 3,9,7,1
2013 △ 4 = more than 503 1
The last digit of 3 ^ 2013 is 3



For any natural number n, when n is odd, add 121, when n is even, divide by 2, this is an operation. Now, if you operate on three digit 241 continuously, will 100 appear in the operation process, why?


241+121=362,362÷2=181;181+121=302,302÷2=151;151+121=272,272÷2=136,136÷2=68,68÷2=34,34÷2=17;17+121=138,138÷2=69;69+121=190,190÷2=95;95+121=216,216÷2=108,108÷2=54,54÷2=27;27+...