3. In this problem, you may use the following formulae without further justification: Σ k²³ = (n(n + ¹)) ². 2 k=1 k=1 n(n+1) 2 3 12 k=1 n(n+1)(2n + 1) k² 1 6 (a) Find a formula in n for each of 71 TL Σ(2k) and Σ(2k – 1). k=1 k=1 (b) Find a formula in n for each of n Σ(2k)² and (2k-1)². k=1 (c) Find a formula in n for each of TL n k=1 TL Σ(2k)³ and (2k-1)³.
3. In this problem, you may use the following formulae without further justification: Σ k²³ = (n(n + ¹)) ². 2 k=1 k=1 n(n+1) 2 3 12 k=1 n(n+1)(2n + 1) k² 1 6 (a) Find a formula in n for each of 71 TL Σ(2k) and Σ(2k – 1). k=1 k=1 (b) Find a formula in n for each of n Σ(2k)² and (2k-1)². k=1 (c) Find a formula in n for each of TL n k=1 TL Σ(2k)³ and (2k-1)³.
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![3. In this problem, you may use the following formulae without further justification:
2
^²³ − ( ²
- - (n(n + ¹)) ².
2
£*
k
k=1
n(n+1)
2
12
k=1
k²
(a) Find a formula in n for each of
(b) Find a formula in n for each of
Σ(2k) and Σ(2k - 1).
k=1
k=1
n
n(n+1)(2n +1)
6
(c) Find a formula in n for each of
n
Σ(2k)² and Σ(2k-1)².
TL
k=1
TL
(d) Only use the formulae above, to compute
Σ(2k)³ and Σ(2k-1)³.
1³22 +33 4²+ +99³-100²,
which is the sum of cubes of all odd integers no greater than 100, minus the sum
of squares of all even integers no greater than 100.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F242e2463-ddfc-4cc7-bb20-37587f636c51%2Ffa5e26ef-d987-46cc-ae9c-d74b2361b7dd%2Fyje8z5s_processed.png&w=3840&q=75)
Transcribed Image Text:3. In this problem, you may use the following formulae without further justification:
2
^²³ − ( ²
- - (n(n + ¹)) ².
2
£*
k
k=1
n(n+1)
2
12
k=1
k²
(a) Find a formula in n for each of
(b) Find a formula in n for each of
Σ(2k) and Σ(2k - 1).
k=1
k=1
n
n(n+1)(2n +1)
6
(c) Find a formula in n for each of
n
Σ(2k)² and Σ(2k-1)².
TL
k=1
TL
(d) Only use the formulae above, to compute
Σ(2k)³ and Σ(2k-1)³.
1³22 +33 4²+ +99³-100²,
which is the sum of cubes of all odd integers no greater than 100, minus the sum
of squares of all even integers no greater than 100.
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