+1124 Hi friends,
just a question...If I look at the sums of each row in Pascal's triangle, I see a ratio of 2. Is it therefore save to say that a Geometric sequence is formed in the triangle?
The following are properties of Pascal's triangle:
1 - The sum of the nth row is 2^n.
2 - The sum of rows 0 through n is 2^(n + 1) - 1.
3 - The generating function of the nth row is (x + 1)^n.
1 - Means the sum of, for example 5th row, ==2^5 ==32
2 - Means the sum of all rows, say from 0 to 5th row, ==2(5 + 1) - 1 ==2^6 - 1==63
3 - Means the generating function of the, say the 5th row, is: (x + 1)^5 ==1 x^5 + 5 x^4 + 10 x^3 + 10 x^2 + 5 x + 1 ==which means the coefficients of the expansion (as you can see) are: 1, 5, 10, 10, 5, 1 - which is the 5th row of Pascal's triangle.
+1124 Thank you guest,
still no idea what all thatmeans, but thank you...do appreciate your time.
+118680
The sum of the 0 row is 2^0 = 1
the sume of the nth row is 2^n
the sum of the 10th row is 2^10 = 1024
the sum of the 20th row is 2^20
etc
+1124 So Melody,
How do we go about finding what type of sequence is generated here?
+118680 it seems to be common pracice to accept that the first line is the 0 line, so it is easiest just to ignore that one
the sequence is 2,4,8,16,32
I can just see that this is 2^n but if that is not immediately obvious to you, you can inspect it as a sequence.
since 4/2 = 8/4 = 16/8 = 32/16 = 2
this is a GP
the first term is 2 and the common difference is 2
\(T_n=ar^{(n-1)}\\~\\ T_n=2*2^{(n-1)}\\~\\ T_n=2^{(n-1+1)}\\~\\ T_n=2^n\)
+1124 Thank you Melody...you really are there for me every time!!!...Stay blessed
