If ∑r=1nTr=64(2n−1)(2n+1)(2n+3)(2n+5), then limn→∞∑r=1n(Tr1) is equal to:
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Step-by-Step Solution
Step 1: Find the general term Tn
We are given the sum of the first n terms, Sn=∑r=1nTr. To find the n -th term Tn, we can subtract the sum of the first n−1 terms, Sn−1, from Sn. This formula allows us to isolate the n -th term.
Step 2: Substitute Sn and Sn−1 and simplify
Substitute the given expression for Sn and Sn−1 into the formula for Tn. After simplifying the terms in Sn−1, we get Sn−1=64(2n−3)(2n−1)(2n+1)(2n+3). We can then factor out common terms.
Step 3: Simplify Tn further
By factoring out the common terms 64(2n−1)(2n+1)(2n+3), we are left with (2n+5) - (2n-3) inside the brackets, which simplifies to 8. This leads to a simplified expression for Tn.
Step 4: Find Tr1 using partial fractions
We need to find the reciprocal of Tr. To sum this series, we use partial fraction decomposition. The expression can be written as a difference of two terms, which is a common technique for telescoping series. We can verify this by combining the terms on the right side.
Step 5: Calculate the sum ∑r=1nTr1
This is a telescoping series. When we write out the terms, most of them will cancel out. For example, the second part of the first term will cancel with the first part of the second term, and so on. Only the first part of the first term and the second part of the last term will remain.
Step 6: Evaluate the telescoping sum
After cancellation, the sum simplifies to 2[1⋅31−(2n+1)(2n+3)1]. This is because all intermediate terms cancel out.
Step 7: Calculate the limit as n→∞
As n approaches infinity, the term (2n+1)(2n+3)1 approaches 0. Therefore, the limit of the sum is 2×31, which simplifies to 32.