Transcription of Solutions to Exercises on Mathematical Induction Math …
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Solutions to Exercises on Mathematical Induction Math …
Solutions to Exercises on Mathematical InductionMath 1210, Instructor: M. Despi cIn Exercises 1-15 use Mathematical Induction to establish the formula forn + 22+ 32+ +n2=n(n+ 1)(2n+ 1)6 Proof:Forn= 1, the statement reduces to 12=1 2 36and is obviously the statement is true forn=k:12+ 22+ 32+ +k2=k(k+ 1)(2k+ 1)6,(1)we will prove that the statement must be true forn=k+ 1:12+ 22+ 32+ + (k+ 1)2=(k+ 1)(k+ 2)(2k+ 3)6.(2)The left-hand side of (2) can be written as12+ 22+ 32+ +k2+ (k+ 1) view of (1), this simplifies to:(12+ 22+ 32+ +k2)+ (k+ 1)2=k(k+ 1)(2k+ 1)6+ (k+ 1)2=k(k+ 1)(2k+ 1) + 6(k+ 1)26=(k+ 1)[k(2k+ 1) + 6(k+ 1)]6=(k+ 1)(2k2+ 7k+ 6)6=(k+ 1)(k+ 2)(2k+ 3) the left-hand side of (2) is equal to the right-hand side of (2). Thisproves the inductive step. Therefore, by the principle of mathematicalinduction, the given statement is true for every positive + 32+ 33+ + 3n=3n+1 32 Proof:Forn= 1, the statement reduces to 3 =32 32and is obviously the statement is true forn=k:3 + 32+ 33+ + 3k=3k+1 32,(3)we will prove that the statement must be true forn=k+ 1:3 + 32+ 33+ + 3k+1=3k+2 32.
Solutions to Exercises on Mathematical Induction Math 1210, Instructor: M. Despi c In Exercises 1-15 use mathematical induction to establish the formula for n 1. 1. 1 2 + 2 2 + 3 2 + + n 2 =
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