WebNov 15, 2024 · Given two positive integers N and K. The task is to evaluate the value of 1 K + 2 K + 3 K + … + N K. Examples: Input: N = 3, K = 4 Output: 98 Explanation: ∑ (x 4) = 1 4 + 2 4 + 3 4, where 1 ≤ x ≤ N ∑ (x 4) = 1 + 16 + 81 ∑ … WebHint: What is $(1+1)^n+(1-1)^n+(1+i)^n+(1-i)^n$? This answer repeatedly takes advantage of one fact, $$g(k)=\\frac{(-1)^k+1}{2}$$ Returns $1$ for a nonnegative e
5.5 Alternating Series - Calculus Volume 2 OpenStax
WebNov 23, 2013 · Subscribe at http://www.youtube.com/kisonecat WebMathematics portal; This article is within the scope of WikiProject Mathematics, a collaborative effort to improve the coverage of mathematics on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks. Mathematics Wikipedia:WikiProject Mathematics Template:WikiProject … gift certificate printable christmas
On the Sums 0 (4k + 1)-n - JSTOR
WebS = Sum from k to n of i, write this sum in two ways, add the equations, and finally divide both sides by 2. We have. S = k + (k+1) + ... + (n-1) + n. S = n + (n-1) + ... + (k+1) + k. When we add these equations, we get 2S on the left side, and n-k+1 column sums that are each n+k on the right side. WebAug 10, 2024 · Please see the Step-by-step explanation for the answers Step-by-step explanation: 1) ∑ 2j The sum of series from j=1 to j=5 is: ∑ = 2 (1) + 2 (2) + 2 (3) + 2 (4) + 2 (5) = 2 + 4 + 6 + 8 + 10 ∑ = 30 2) This question is not given clearly so i assume the following series that will give you an idea how to solve this: ∑ 2k² Web1. Sum of (2k-1) (k^2-1)/ (k+1) ( (k^2+4)^2) with k=1 -> infinite 2. Sum of sqrt (n)/ (n-1) with n=2 -> infinite 3. Sum of 1/ (2n+3) with n=1 -> infinite 4. Sum of (n+2)/ ( (n+1)^3) with n=3 -> infinite Expert's answer Download Answer Need a fast expert's response? Submit order and get a quick answer at the best price fry kale recipe