MA132 Foundations Exercises I Solutions - Of the Clux

1MA132 Foundations Exercises I Solutions1. Use induction to prove the following formulae:(i) 1 2 + 2 2 + · · · + n 2 = 1 n(n + 1)(2n + 1) (ii) 1 + 3 + 5 + 7 + · · · + (2n + 1) = (n + 1)262. (Taken from T.M.Apostol, Calculus, Volume 1)(i) Observe that1 = 11 − 4 = −(1 + 2)1 − 4 + 9 = 1 + 2 + 31 − 4 + 9 − 16 = −(1 + 2 + 3 + 4).Guess the general law and prove it by induction.Solution: The general law isn∑(−1) k−1 k 2 = (−1) n−1k=1n∑k.The formulae given show that this is true for n = 1, 2, 3 and 4. Suppose it true for n. Thenk=1∑n+1(−1) k−1 k 2 =k=1n∑(−1) k−1 k 2 + (−1) n (n + 1) 2k=1= (−1) n−1 n∑k=1k + (−1) n (n + 1) 2 = (−1) n( (n + 1) 2 − 1 n(n + 1))2= (−1) n (n + 1)(n + 1 − n 2 ) = (−1)n (n + 1)( n + 22∑n+1) = (−1) n k.So it’s true for n + 1 as well. This, together with the initial step, proves that it holds for alln ≥ 1.(ii) Observe that1 − 1 2= 1 2(1 − 1 2 ) (1 − 1 3 ) = 1 3(1 − 1 2 ) (1 − 1 3 ) (1 − 1 4 ) = 1 4Guess the general law and prove it by induction.Solution: The general law isk=1(1 − 1 2 )(1 − 1 3 )· · ·(1 − 1 n ) = 1 nfor n ≥ 2. The formulae displayed show that it is true for n = 2, 3, 4. Suppose it true for n.Then(1 − 1 1)· · ·(1 −2 n + 1 ) = (1 − 1 2 )· · ·(1 − 1 n )(1 − 1n + 1 )

= 1 n (1 − 1n + 1 ) = 1 n × nn + 1 = 1n + 1 .So it holds for n + 1. hence by induction it’s true for all n ≥ 2.(iii) Guess a general law which simplifies the productand prove it by induction.( 1)( 1)( 1 1 )1 − 1 − 1 −)· · ·(1 −4 9 16 n 2Solution: We have ( )1 −14 =34( )( )1 −14 1 −19 =3× 8 = 2 4 9 3( )( )( )1 −14 1 −19 1 −116 =2× 15 = 5 3 16 8There doesn’t seem to be an obvious pattern until we realise that 2 = 4 ; as soon as we do,3 6the obvious guess is( 1)( 1 1 ) n + 11 − 1 −)· · ·(1 − =4 9 n 2 2n .This holds for n = 2, 3, 4. Suppose it is true for n. Then( 11 ) n + 11 −)· · ·(1 − =4 (n + 1) 2 2n× ( 1 )1 −(n + 1) 2= n + 12n × (n + 1)2 − 1= n + 1(n + 1) 2 2n× n2 + 2n(n + 1) = n + 22 2(n + 1) .Thus by induction it holds for all n.3. (i) Prove that for all n ∈ N, 10 n − 1 is divisible by 9.(ii) There is a well-known rule that a natural number is divisible by 3 if and only if the sumof the digits in its decimal representation is divisible by 3. Part (i) of this exercise is thefirst step in a proof of this fact. Can you complete the proof? Hint: the point is to comparethe two numbers k 0 + k 1 × 10 + k 2 × 10 2 + · · · + k n × 10 n and k 0 + k 1 + · · · + k n .(iii) Does your proof enable you to say something stronger than the rule stated?(iv) Is there a similar rule for divisibility by 9?Solution:(i)10 − 1 = 9,10 2 − 1 = 99 = 9 × 10 + 910 3 − 1 = 999 = 9 × 10 2 + 9 × 10 + 9,so the pattern is obvious.(ii) The number N represented in the decimal notation as k n k n−1· · ·k 1 k 0 is equal to k 0 +k 1 ×10 + · · · + k n × 10 n . The difference between N and the sum of its digits, S, isk 0 + k 1 × 10 + · · · + k n × 10 n − (k 0 + k 1 + · · · + k n )= k 1 × (10 − 1) + k 2 × (10 2 − 1) + · · · + k n × (10 n − 1).