Question

If

A=1+1/[1+2]+1/[1+2+3]+.....+1/[1+2+3+....1000]

and
B=1+[3/4]+[9/(16×2!)]+[27/(64×3!)]+...+[3⁹⁹/(4⁹⁹×99!)]+...if the value of "B" converges.

Then find [A²+B²]?

How do I do this? I need details solution plz put me through......

Answer 1

the A series is relatively straightforward:
A=1+1/[1+2]+1/[1+2+3]+.....+1/[1+2+3+....1000]
= 1 + 1/2 + 1/6 + 1/10 + 1/15 + .... for 1000 terms

sum(1) = 1
sum(2) = 1 + 1/2 = 4/3
sum(3) = 4/3 + 1/6 = 3/2 or 6/4
sum(4) = 3/2 + 1/10 = 8/5
sum(5) = 8/5 + 1/15 = 5/3 or 10/6
looking at the last fraction in each case, my conjecture is that
sum(n) = 2n/(n+1)

then A = sum(1000) = 2000/1001

B is another story.
B = 1 + (3/4)/1! + (3/4)^2 /2! + (3/4)^3 /3! + (3/4)^4 /4! + .... + (3/4)^99 / 99!
= 1 + 3/4 + 9/32 + 27/384 + 81/6144 + ...

sum(1) = 1
sum(2) = 1 + 3/4 = 7/4
sum(3) = 7/4 + 9/32 = 65/32
sum(4) = 65/32 + 27/384 = 269/128
sum(5) = 269/128 +81/6144 = 4331/2048
.....
So far I have noticed that the denominators are all powers of 2
2^0, 2^2, 2^5, 2^7, 2^11 , ....
for the numerators of 1, 7, 65, 269, 4331, ... I have not found a pattern yet.

however , B does converge and using some calculations on my scientific calculator
I found
Sum(7) = 2.116971
sum(8) = 2.116997 <------ converging to appr 2.117

(A^2 + B^2) = ( (2000/1001)^2 + 2.117^2) = appr 8.4737

Don't know if this helps, but it was sort of fun

Answer 2

Thank you reiny am so greatful

I understand it