1.n=99。

Fa 1

Note that the left end is a difference sequence, which can be subtracted by classical dislocations:

Remember s = I+2i 2+3i 3+…+ni n,

is = i^2+2i^3+…+(n- 1)i^n+ni^(n+ 1)

Two types of subtraction:

( 1-i)s=i+i^2+i^3+…+i^(n- 1)+i^n-ni^(n+ 1)

At the right end of the above formula, except that the last term is geometric series, you can sum up and find the expression of S.

Then let S=-50( 1+i).

Method 2 (simpler method)

The period of sequence I n is 4, and the first term is: i,-1, -i, 1. The real part and imaginary part can be discussed separately. Remember that the left end of the original form is S.

The real part of s consists of even terms in I n, let t be the largest even number ≤n, and RES = 2i 2+4i 4+…+Ti t =-2+4-6+8+…+(-2) (t/2+1).

It is easy to see that when there is an even addend, this sum is positive, and when there is an odd addend, this sum is negative. ReS=-50 is known, so there is an odd addend in the above formula. After pairing, it can be regarded as the addition of several -2, thus finding t=98.

Similarly, let s be the largest odd number ≤n, and after comparing the imaginary part, we can get s=99.

So n=99.

2.5/2

All positive factors of10 n can be written as 2 s * 5 t, where s and t are natural numbers ≤ n.

a(n)=σ(t = 0，n)σ(s = 0，n) {2^s*5^t}

=σ(t=0,n){5^t*σ(s=0,n)2^s}

=[2^(n+ 1)- 1]σ(t=0,n){5^t}

=[2^(n+ 1)- 1][5^(n+ 1)- 1]/4=(5/2)* 10^n+…

The latter part … is divided by10 n, and the limit is 0. So the limit is 5/2.

3.496

The original formula is the multiplication of three sums, and the power of each sum contribution in each multiplied x198 is considered. This is equivalent to finding the number of groups (a, b, c) of solutions of the following equations:

A+b+c = 198, a, b and c are natural numbers, and a ≤ 30; b，c≤99。

In the above range, for a fixed A and B, we can take 99, 98, …99-a, and each B corresponds to a range of C, so after fixing A, there are a+ 1 paths to take (b, c) * * to satisfy the equation.

So the number of groups of solutions of the above formula = 1+2+…3 1=496.

4. 1/4

For simplicity, let's assume that surface A is 1, surface B is 6, and the probability of turning to surface A is X. 。

The sum of squares is 7, and there are three possible combinations (3+4 or 4+3, 2+5 or 5+2, 1+6 or 6+ 1).

The sum of the probabilities of the first two cases = (1/6) 2 * 2 * 2 = 1/9.

Sum of probabilities in the third case =2x( 1-x).

1/9+2x (1-x) =11/72, and x >；; 1/6，x= 1/4。