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.
How do i find [A²+B²]?
A = 2000/1001
B ≈ 2.117
But how sir obleck?
To find the value of [A²+B²], we need to calculate the values of A and B first, and then square both of them.
Let's start with calculating the value of A.
A = 1 + 1/[1+2] + 1/[1+2+3] + ... + 1/[1+2+3+....+1000]
We can see that the denominator of each fraction is increasing by 1 each time. So, we can rewrite each term as follows:
A = 1 + 1/3 + 1/6 + ... + 1/(sum of numbers from 1 to 1000)
Now, let's find the sum of numbers from 1 to 1000. We can use the formula for the sum of an arithmetic series:
Sum = (n/2) * (first term + last term)
In this case, the first term is 1 and the last term is 1000. So,
Sum = (1000/2) * (1 + 1000)
= 500 * 1001
= 500500
Substituting this value back into A, we get:
A = 1 + 1/3 + 1/6 + ... + 1/500500
Now, let's move on to calculating the value of B.
B = 1 + [3/4] + [9/(16×2!)] + [27/(64×3!)] + ... + [3⁹⁹/(4⁹⁹×99!)] + ...
We can see that this is a geometric series where each term is obtained by multiplying the previous term by a common ratio of 3/4. The first term is 1, so the formula to find the sum of an infinite geometric series can be used:
Sum = (first term) / (1 - common ratio)
In this case, the first term is 1 and the common ratio is 3/4. So,
Sum = 1 / (1 - 3/4)
= 1 / (1/4)
= 4
Now, we have the values of A and B:
A = 1 + 1/3 + 1/6 + ... + 1/500500
B = 4
To find [A²+B²], square both A and B and add them together:
[A²+B²] = (A * A) + (B * B)
Now, substitute the values of A and B into the equation and calculate the result.