Math

Try this

Sum of odds is equal to square
Proof by Induction, show the first case is true (n=1), show the nth case is true, then show by extension that the (n+1)th case is true.
For i=1 to n Σ [2(i-1)+1]=n^2
For i=1 to (n+1) Σ [2(i-1)+1]=(n+1)^2=n^2+2n+1
And
{for i=1 to (n+1)} Σ [2(i-1)+1]= {for i=1 to n} Σ [2(i-1)+1] +[2(n+1-1)+1]
{for i=1 to n} Σ [2(i-1)+1] +[2(n+1-1)+1]={for i=1 to n} Σ [2(i-1)+1] +[2(n)+1]=n^2+2n+1
so its true for the (n+1)th case

Pascal's Triangle
1) 7 elements in the 6th row
2) n+1 elements in the nth row
3) sums are 2,4,8,16,32,64,….,2^n
4) They double
5) 2^8
6) Binary base
7) 2^n
8) (a+b)(a+b)=a^2+2ab+b^2
9) (a+b)(a+b)(a+b)=( a^2+2ab+b^2)(a+b)=a^3+3a^2b+3ab^2+b^3
10) expansion of (a+b)^n
11) a^3+3a^2b+3ab^2+b^3

Look the first series problem
1) 6th row=2*6+1 =13 elements
2) First number is one more than the last number in the 5th row which would be 5^2=25, or the first number is (6-1)^2+1=26
3) nth row has 2n+1 elements
4) Last element in the nth row is n^2
5) First element in the nth row is one more than the (n-1)th row which is (n-1)^2. So the first element is (n-1)^2+1=n^2-2n+1+1=n^2-2n+2

Rap c∫;?/

Thanks alot raprap. Question 1 was really confusing. But thanks for the help.

What are you studying by the way?