G1,G2,G3 and common ratio?

Question

Let G1,G2,G3,.....Gp denote the respective sums to infinity of p geometric progressions,with first terms 1,2,3,...p and common ratios 1/2,1/3,.........1/(p+1) respectively...whats the value of G1+G2+G3+...Gp ?


Question:

"..Let G1,G2,G3,.....Gp denote the respective sums to infinity of p geometric progressions..."


does this statements means below ?

so, G1=1
G2=1+2
G3=1+2+3

Gp=1+2+3.....+p




Question:

"...and common ratios 1/2,1/3,.........1/(p+1)..."

what does it mean ? common ratios of what ? common ratio's of G1,G2,G3 or what ?...

please explain

Answer 1

Answer: p *(p+3)/2

Worked Out Solution:

Sum to infinity of a GP= a/(1-r) where a= 1st Term and r=common ratio;

Therefore, sum of 1st infinite progression = G1=1-(1/2))= 2
Therefore, sum of 2nd infinite progression = G2= 1-(1/3))=3
Therefore, sum of pth infinite progression = G3=p/(1-(1/(p+1)))= p+1

Therefore, G1 +G2+.....Gp = 2+3+4+....(p+1) = p*(p+3)/2

Answer 2

G1 = 1 + 1/2 + 1/4 + 1/8 + 1/16 + ...
G2 = 2 + 2/3 + 2/9 + 2/27 + 2/81 + ...
G3 = 3 + 3/4 + 3/16 + 3/64 + 3/256 + ...
...
Gp = p + p/(p+1) + p/(p+1)^2 + ...

Each line is a "sum to infinity of a geometric progression", and the terms of the sum are obtained by multiplying the preceding term by the "common ratio".

The question is now: how much is the some of the first p rows like this? First, it is handy to know that

b + ba + ba^2 + ba^3 + ... = ab / (1 - a)

(b is the common term, a is the common ratio of a geometric progression). For row Gn, we have b = n and a = 1/(n+1), so

Gn = n/(n+1) / [1 - 1/(n+1)] = n/(n+1) / [(n + 1 - 1)/(n+1)] = n/n = 1

All sums are equal to ONE! Therefore, adding the first p rows is simply adding 1 + 1 + 1 + ..., p times; the answer should be p.

Answer 3

for gi sum, si=i/1-(1/i+1) [ sum of an infinite gp is a/1-r
where a is the starting term and r is common ratio]
=i*(i+1)/i
=i+1
therefore the reqd sum is sum of all si's

= n*(n-1)/2 + n
= (n^2 - n + 2n)/2
= (n^2 +n)/2

Answer 4

trouble-free ratio is the consistent nonzero genuine huge variety it fairly is accelerated to the former time period to obtain each and every time period after the first. meaning: first time period is 4. 2d time period is 4 x (3) = 12 0.33 time period is 12 x (3) = 36 fourth time period is 36 x (3) = 108 for this reason, the easy ratio is 3#. desire you recognize what i recommend... :S