\sum_(n=0)^5 (2n-n^(2))
Added by Kourtnie L.
Step 1
For n = 0: 2n - n^2 = 2(0) - 0^2 = 0 - 0 = 0 For n = 1: 2n - n^2 = 2(1) - 1^2 = 2 - 1 = 1 For n = 2: 2n - n^2 = 2(2) - 2^2 = 4 - 4 = 0 For n = 3: 2n - n^2 = 2(3) - 3^2 = 6 - 9 = -3 For n = 4: 2n - n^2 = 2(4) - 4^2 = 8 - 16 = -8 For n = 5: 2n - n^2 = 2(5) - Show more…
Show all steps
Close
Your feedback will help us improve your experience
Zhumagali Shomanov and 56 other Calculus 1 / AB educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
\small \textup{What is the value of} \sum_{n=1}^{5}\left (2n-3 \right )?
Benjamin C.
$\sum_{n=0}^{4} n !$
Sequences, Series, and Probability
Series
. $\sum_{n=0}^{\infty} \frac{5^{\pi}(n !)^{2}}{(2 n) !}$
INFINITE SERIES, POWER SERIES
Tests for convergence of series of positive terms; absolute convergence
Recommended Textbooks
Calculus: Early Transcendentals
Thomas Calculus
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD