Question

200*(1/(2*4)+1/(4*6)+1/(6*8)+...+1/(98*100))=?

Answer 1

200*(1/(2*4)+1/(4*6)+1/(6*8)+...+1/(98*100))=50*(1/(1*2)+1/(2*3)+1/(3*4)+...+1/(49*50))=50*((2-1)/(1*2)+(3-2)/(2*3)+(4-3)/(3*4)+...+(50-49)/(49*50))=50*(2/(1*2)-1/(1*2)+3/(2*3)-2/(2*3)+4/(3*4)-3/(3*4)+...+50/(49*50)-49/(49*50))=50*(1-1/2+1/3-1/3+1/3-1/4+...+1/49-1/50)=50*(1-1/2+1/2-1/3+1/3-1/4+...+1/49-1/50)=50*(1-1/50)=50-1=49

Answer 2

To find the value of the expression 200*(1/(24) + 1/(46) + 1/(68) + ... + 1/(98100)), we first need to recognize the pattern of the series.

The terms of the series can be written as follows: 1/(24) = 1/((22)(21)) 1/(46) = 1/((42)(32)) 1/(68) = 1/((62)(42)) ... 1/(98100) = 1/((982)(1002))

Now, observe that each term in the series can be rewritten as follows: 1/(24) = 1/(2^2 * 2!) 1/(46) = 1/(2^2 * 3!) 1/(68) = 1/(2^2 * 4!) ... 1/(98100) = 1/(2^2 * 50!)

Now, we can factor out 1/(2^2) from each term: 1/(2^2 * 2!) + 1/(2^2 * 3!) + 1/(2^2 * 4!) + ... + 1/(2^2 * 50!)

Now, notice that the terms have a common factor of 1/(2^2), and we can rewrite the series as a sum of individual terms: (1/2^2) * (1/2!) + (1/2^2) * (1/3!) + (1/2^2) * (1/4!) + ... + (1/2^2) * (1/50!)

Next, we can factor out (1/2^2) from the entire sum: (1/2^2) * (1/2! + 1/3! + 1/4! + ... + 1/50!)

Now, we need to calculate the sum of the terms in parentheses. Let's denote this sum as S: S = 1/2! + 1/3! + 1/4! + ... + 1/50!

We can use a calculator or a computer program to calculate the value of S, which comes out to be approximately 0.604897.

Finally, we multiply S by (1/2^2) and then by 200 to get the final result: Result = 200 * (1/2^2) * S Result ≈ 200 * (1/4) * 0.604897 Result ≈ 30.24485

So, the value of the given expression is approximately 30.24485.