I have some math problems. Someone please help!?

Question

1. What is the remainder when 1+x^1111+x^2222+.............x^7777+x^8888+x^9999 is divided by 1+x^1+x^2+.............x^7+x^8+x^9?
2, Which one is greater 99^n +100^n or 101^n? given that n>50 and n is an integer.
3. A series is given such that
pi=3*3^(1/2)+24(1/12-1/(5*2^5)-1/(28*2^7)-1/(72*2^9)-..............)
how many term is needed to obtain the value of pi to sixteen decimal place?
4. When x^1951-1 is divided by x^4+x^3+2x^2+x+1 what is the co-efficient of x^14 ?
5. Another series of pi
pi=6(1/2+1/(2*3*2^3)+1*3/(2*4*5*2^5)+.........)
how many term is needed to obtain the value of pi to ten decimal place?

Some letter missed
Q#3 pi=3*3^(1/2)+24{1/12-1/(5*2^5)-1/(28*2^7)-1/(72*2^9)-....}
Q#5
pi=6{1/2+1/(2*3*2^3)+1*3/(2*4*5*2^5).....}

Thnx in anticipation. :)

I must appreciate steiner1745 for his attempt. Get going oops!

Answer 1

I'm working on these!
The first question is missing an exponent on the last x.
Should it be x^8888?
Let's do the second question.
The answer is that 101^n is greater for n > 50.
Actually, the result is also true for n = 49 and 50.
Now direct computation with PARI shows that
the result is true for 49 ≤ n ≤ 100,
so it is enough to prove it for n > 100.
To see this, note that 99^n + 100^n < 100^n + 100^n = 2*100^n
for all n.
Now let's look at 101^n. We have
101^n = (100+1)^n > 100^n + n*100^(n-1).
Suppose now n > 100.Then
100^n + n*100^(n-1) > 100^n + 100*100^(n-1) = 2*100^n
> 99^n + 100^n,
so the result follows.

No. 1: Some experiments
with PARI show that the remainder is 0.
Here is the proof: I will assume that the missing
exponent is 8888.
Let g(x) = 1+x^1+x^2+...+ x^7+x^8
and f(x) = 1+x^1111+x^2222+...+ x^7777 + x^8888.
It is enough to show that every root of g is a root of f.
For then every factor of g is a factor of f
and that implies g divides f.
Now g also = (x^9-1)/(x-1)
so every root w of g is a ninth root of unity(w <>1)
and 1 + w + w^2 + w^3 + w^4 + w^5 + w^6 + w^7 + w^8 = 0 (1)
Now let's look at f(w).
We get 1 + w^1111 + w^2222 + w^3333 + w^4444
+ w^5555 + w^6666 + w^7777 + w^8888.
Since w is a 9th root of unity, we can reduce all
the exponents modulo 9.
For example w^1111 = w^1107*w^4 = w^4,
since 1107 = 9*123.
So, f(w) = 1 + w^4 + w^8 + w^3 + w^7 + w^2
+ w^6 + w + w^5,
which is (1), with the summands in different order.
Since w is a 9th root of unity (w <>1), that
means f(w) = 0 and we are done.

Answer 2

the answer is 4 # 1. 84
#2.293
#3.83
#4.9
#5.92736

Answer 3

48?