Question

please solve this question

Answer 1

Answer:

Step-by-step explanation:

What is the value of the series {(3^2-1) / (3^2+1) + (4^2-1) / (4^2+1) + (5^2-1) / (5^2+1) + _ _ _ _ _ _ _ _ (99^2-1) / (99^2+1)}?

Solution:

While this is not an exact answer, it is ,I believe, a good approximation.

We have the following:

∑99x=3x2−1x2+1

=∑99x=3x2−1+2−2x2+1

=∑99x=3x2+1x2+1−2x2+1

=∑99x=31−2x2+1

=∑99x=31−2∑99x=31x2+1

=97−2∑99x=31x2+1

Now, I’m not sure how you would evaluate that summation, but as an approximation you can integrate it. This gives:

≈97−2∫993dx1+x2

≈97−2(arctan99−arctan3)

≈97−2(0.31164)

≈97−0.623

≈96.377