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MAT2612 ASSIGNMENT 1 2021
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This document contains MAT2612 ASSIGNMENT 1 2021 solutions. All workings are shown clearly step by step and explanations are also provided.
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MAT2612ASSIGNMENT 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
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