MAT2612 ASSIGNMENT 1 2021

MAT2612

ASSIGNMENT 1 2021

QUESTION 1


a).


(2𝑛 + 1)2
1 + 2 + 3+. . . . . . . . . +𝑛 ≤
8

𝐵𝑎𝑠𝑖𝑠 𝑠𝑡𝑒𝑝: 𝑃𝑟𝑜𝑣𝑒 𝑖𝑡𝑠 𝑡𝑟𝑢𝑒 𝑓𝑜𝑟 𝑛 = 1

(2(1) + 1)2
1≤
8

(3)2
1≤
8
9
1≤ ℎ𝑜𝑙𝑑𝑠 𝑓𝑜𝑟 𝑛 = 1
8



𝐴𝑠𝑠𝑢𝑚𝑒 𝑖𝑡𝑠 𝑡𝑟𝑢𝑒 𝑓𝑜𝑟 𝑛 = 𝑘

(2𝑘 + 1)2
1 + 2 + 3+. . . . . . . . . +𝑘 ≤
8



𝑃𝑟𝑜𝑣𝑒 𝑓𝑜𝑟 𝑛 = 𝑘 + 1

𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑎𝑠𝑠𝑢𝑚𝑝𝑡𝑖𝑜𝑛:

(2𝑘 + 1)2
1 + 2 + 3+. . . . . . . . . +𝑘 ≤
8

𝐴𝑑𝑑 (𝑘 + 1) 𝑡𝑜 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒𝑠

(2𝑘 + 1)2
1 + 2 + 3+. . . . . . . . . +𝑘 + (𝑘 + 1) ≤ + (𝑘 + 1)
8

4𝑘 2 + 4𝑘 + 1
≤ + (𝑘 + 1)
8

, 4𝑘 2 + 4𝑘 + 1 8(𝑘 + 1)
≤ +
8 8

(4𝑘 2 + 4𝑘 + 1) + 8(𝑘 + 1)
≤
8

4𝑘 2 + 4𝑘 + 1 + 8𝑘 + 8
≤
8

4𝑘 2 + 12𝑘 + 9
≤
8
(2𝑘 + 3)(2𝑘 + 3)
≤
8

(2𝑘 + 3)2
≤
8

(2𝑘 + 2 + 1)2
≤
8

(2(𝑘 + 1) + 1)2
≤ ℎ𝑜𝑙𝑑𝑠 𝑓𝑜𝑟 𝑛 = 𝑘 + 1
8



(2𝑛 + 1)2
1 + 2 + 3+. . . . . . . . . +𝑛 ≤ 𝑖𝑠 𝑡𝑟𝑢𝑒 𝑏𝑦 𝑚𝑎𝑡ℎ𝑒𝑚𝑎𝑡𝑖𝑐𝑎𝑙 𝑖𝑛𝑑𝑢𝑐𝑡𝑖𝑜𝑛
8



b).


𝑛2 − 3𝑛 + 4 𝑖𝑠 𝑒𝑣𝑒𝑛 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑛 ≥ 1



𝐵𝑎𝑠𝑖𝑠 𝑠𝑡𝑒𝑝: 𝑃𝑟𝑜𝑣𝑒 𝑖𝑡𝑠 𝑡𝑟𝑢𝑒 𝑓𝑜𝑟 𝑛 = 1

𝑛2 − 3𝑛 + 4 = (1)2 − 3(1) + 4

=1−3+4

= 2 (𝑒𝑣𝑒𝑛) ℎ𝑜𝑙𝑑𝑠 𝑓𝑜𝑟 𝑛 = 1.



𝐴𝑠𝑠𝑢𝑚𝑒 𝑖𝑡𝑠 𝑡𝑟𝑢𝑒 𝑓𝑜𝑟 𝑛 = 𝑘

𝑛2 − 3𝑛 + 4 = 𝑘 2 − 3𝑘 + 4

𝐿𝑒𝑡: 𝑘 2 − 3𝑘 + 4 = 2𝑚 , 𝑤ℎ𝑒𝑟𝑒 𝑚 𝑖𝑠 𝑎𝑛 𝑖𝑛𝑡𝑒𝑔𝑒𝑟