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......
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
Thank you reiny am so greatful
I understand it
To find the value of [A² + B²], we need to calculate the values of A and B separately, and then substitute them into the equation.
1. Calculating A:
Let's break down the expression for A and find its value step by step.
A = 1 + 1/(1+2) + 1/(1+2+3) + ... + 1/(1+2+3+...+1000)
To simplify the expression, we'll start by considering each term separately:
- The first term is simply 1.
- The second term is 1/(1+2) = 1/3.
- The third term is 1/(1+2+3) = 1/6.
- The fourth term is 1/(1+2+3+4) = 1/10.
We can observe a pattern here. Each term has the form 1/(1+2+3+...+n). So, in general:
- The nth term is 1/(1+2+3+...+n) = 1/(n(n+1)/2).
Now, let's write out the expression for A using this general form:
A = 1 + 1/3 + 1/6 + 1/10 + ...
As we go forward, we can rewrite the terms in A as fractions with a common denominator:
A = (6/6) + (2/6) + (1/6) + (3/30) + ...
Simplifying this expression, we get:
A = 12/6 + 3/6 + 1/6 + 3/30 + ...
Combining the terms, we have:
A = (12+3+1)/6 + 3/30 + ...
Simplifying further:
A = 16/6 + 1/10 + ...
We can see that the terms beyond the first two terms (16/6 and 1/10) form a geometric series.
The sum of an infinite geometric series with a common ratio less than 1 is given by the formula: a / (1 - r), where 'a' is the first term and 'r' is the common ratio.
In this case, the first term is 1/10 and the common ratio is 1/10.
Using the formula, the sum of the geometric series is:
Sum = (1/10) / (1 - 1/10) = (1/10) / (9/10) = 1/9.
So, the value of A is:
A = 16/6 + 1/9 = (8/3) + (1/9) = 25/9.
Hence, A = 25/9.
2. Calculating B:
Let's break down the expression for B and find its value step by step.
B = 1 + 3/4 + 9/(16×2!) + 27/(64×3!) + ... + (3⁹⁹)/(4⁹⁹×99!) + ...
Here, it is evident that the terms form a pattern: each term is of the form (3^n)/((4^n) × n!).
To calculate B, we need to determine whether the series converges or not. This can be done by checking the common ratio sequence.
The common ratio between consecutive terms can be found by dividing the (n+1)th term by the nth term:
(3^(n+1))/((4^(n+1)) × (n+1)!) ÷ (3^n)/((4^n) × n!)
Simplifying this, we get:
((3^(n+1))/((4^(n+1)) × (n+1)!)) × ((4^n) × n!)/(3^n)
This can be further simplified to:
3/4 × 1/(n+1)
As we can see, the common ratio is 3/4 × 1/(n+1). Since this ratio approaches zero as n approaches infinity, we can conclude that the series converges.
Now, to find the value of B, we need to calculate the sum of this converging series.
Let's write out the expression for B again:
B = 1 + 3/4 + 9/(16×2!) + 27/(64×3!) + ... + (3⁹⁹)/(4⁹⁹×99!) + ...
The general form of the nth term can be written as:
((3^n)/((4^n) × n!))
To find the sum of the series, we can use the formula for the sum of a converging geometric series:
Sum = (a × r) / (1 - r), where 'a' is the first term and 'r' is the common ratio.
Here, the first term is 1, and the common ratio is 3/4 × 1/(n+1). As we previously determined, the common ratio converges to zero.
Using the formula, the sum of the series is:
Sum = (1 × 0) / (1 - 0) = 0 / 1 = 0.
Hence, B = 0.
3. Calculating [A² + B²]:
Now that we have found the values of A and B, we can substitute them into the equation [A² + B²]:
[A² + B²] = [(25/9)² + 0²]
Calculating the square of 25/9:
[(25/9)² = (25²)/(9²) = 625/81.
Substituting the values:
[(25/9)² + 0²] = (625/81) + 0 = 625/81.
Therefore, [A² + B²] = 625/81.