5760 Consider the sum4k=1[(k + 1)3 k3]=[53 43]+[43 33]+ [33 23]+[23 13]= 53 13 = 124For
Chapter 5, Problem 58(choose chapter or problem)
Consider the sum
\(\sum_{k=1}^{4}\left[(k+1)^{3}-k^{3}\right]=\left[5^{3}-4^{3}\right]+\left[4^{3}-3^{3}\right]+\left[3^{3}-2^{3}\right]+\left[2^{3}-1^{3}\right]=5^{3}-1^{3}=124\)
For convenience, the terms are listed in reverse order. Note how cancellation allows the entire sum to collapse like a telescope. A sum is said to telescope when part of each term cancels part of an adjacent term, leaving only portions of the first and last terms uncanceled. Evaluate the telescoping sums in these exercises.
\(\sum_{k=1}^{50}\left(\frac{1}{k}-\frac{1}{k+1}\right)\)
Equation Transcription:
Text Transcription:
sum_k=1 ^4 [(k+1)^3 -k^3 ]=[5^3 -4^3 ]+[4^3 -3^3 ]+[3^3 -2^3]+[2^3 -1^3 ]=5^3 -1^3 =124
sum_k=1 ^50 (1/k - 1/k+1 )
Unfortunately, we don't have that question answered yet. But you can get it answered in just 5 hours by Logging in or Becoming a subscriber.
Becoming a subscriber
Or look for another answer