Tiêu đề MATHEMATICAL INDUCTION (Q)-1.pdf ✅

\\ Chapter – 1 The word induction means the method of reasoning about a general statement from the conclusion of particular cases. Inductions starts with observations. It may be true but then it must be so proved by the process of reasoning. Else it may be false but then it must be shown by finding a counter example where the conjecture fails. In mathematics there are some results or statements that are formulated in terms of n, where n N . To prove such statements we use a well suited method, based on the specific technique, which in known as principle of mathematical induction. A statement which is either true or false is called a proposition or statement. Pn denotes a proposition whose truth value depends on natural variable ‘n’. For example 2 2 2 2 1  2  3  ......n =    6 n n  1 2n  1 is a proposition whose truth value depends on natural number n. We write      6 1 2 1 : 1 2 3 ....... 2 2 2 2       n n n P n n , where P(4) means     6 4 4 1 8 1 1 2 3 4 2 2 2 2       To prove the truth of proposition Pn depending on natural variable n, we use mathematical induction. The statement Pn is true for all n N if (i) P(1) is true. (ii) P(m) is true  Pm  1 is true. The above statement can be generalized as Pn is true for all n N and n  k if (i) Pk is true. (ii) Pm is true m  k  Pm  1 is true. WORKING RULE To prove any statement Pn to be true for all n  k with the help of first principle of mathematical induction we follow the following procedure: Step (i) Check if the statement is true or false for n = k. Step (ii) Assume the statement is true for n = m. Step (iii) Prove the statement is true for n  m  1. Illustration 1 INTRODUCTION 1 THEORY CONTENT OF MATHEMATICAL INDUCTION PROPOSITION 2 3 FIRST PRINCIPLE OF MATHEMATICAL INDUCTION APPLICATION OF FIRST PRINCIPLE OF MATHEMATICAL INDUCTION 4
Question: Prove by the principle of mathematical induction that for all n  N :  1 1 1 ...... 3.4 1 2.3 1 1.2 1        n n n n . Solution: Let Pn be the statement given by Pn :  1 1 1 ...... 3.4 1 2.3 1 1.2 1        n n n n Step I: We have, P1 : 1 1 1 1.2 1   Since, 2 1 1 1 1 1.2 1    So, P(1) is true. Step II: Let Pm be true, then  1 1 1 ...... 3.4 1 2.3 1 1.2 1        m m m m (i) We shall now show that Pm 1 is true. If P(m) is true. For this we have to show that       1 1 1 1 1 1 1 1 1 ...... 3.4 1 2.3 1 1.2 1              m m m m m m Now,    1  1 1 1 1 1 ...... 3.4 1 2.3 1 1.2 1         m m m m =    1  1 1 1 1 1 ...... 3.4 1 2.3 1 1.2 1               m m m m =  1 1 1 1 1     m  m m m [using (i)] =            1 2 1 2 2 1 1 1 2 1 1 1 1 2 2                   m m m m m m m m m m = 2 1   m m  Pm 1 is true. Thus, Pm is true  Pm 1 is true. Hence, by the principle of mathematical induction, the given statement is true for all n  N . Illustration 2 Question: For every positive integer n, prove that n n 7  3 is divisible by 4. Solution: We have   n n P n : 7  3 is divisible by 4 We note that 1: 7 3 4 1 1 P   , which is divisible by 4. Thus Pn is true for n = 1 Let Pk  be true for some natural number k. i.e.,   k k P k : 7  3 is divisible by 4. We get k k 7  3 = 4d, where d  N ...(i) Now, we wish to prove that Pk  1 is true whenever Pk  is true. i.e., we have show m k k 7 3 4 1 1     Now,  1  1  1  1 7 3 7 7.3 7 3 3          k k k k k k =         k k k k 7 7  3  7  3 3  7 4d  7  3 3 =   k 7 4d  4.3 [using (i)]   k  4 7d  3 = 4m From the last line, we see that  1  1 7 3    k k is divisible by 4. Thus, Pk  1 is true when Pk  is true.
Therefore, by principle of mathematical induction the statement is true for every positive integer n. Illustration 3 Question: Prove that 2.7  3.5  5 n n is divisible by 24 for all n  N . Solution: Let the statement Pn be defined as  : 2.7  3.5  5 n n P n is divisible by 24. Now, Pn is true for n = 1, since 2.7  3.5  5  24 , which is divisible by 24. Assume that Pk  is true. i.e. q k k 2.7  3.5  5  24 , when q  N ...(i) Now, we wish to prove that Pk  1 is true whenever Pk  is true. We have, 2.7 3.5 5 2.7 .7 3.5 .5 5 1 1 1 1      k k k k = 7[2.7  3.5  5  3.5  5]  3.5 .5  5 k k k k = 7[24  3.5  5] 15.5  5 k k q [using (i)] = 7 24  21.5  35 15.5  5 k k q = 7 24  6.5  30 k q = 7 24 30 5 1 1    k  q =7  24q  30 4p [since 5 1 1  k  is a multiple of 4 as n n x  y is divisible by x = y] = 7  24q  120p = 247q  5p = 24  r, r  7q  p , is some natural number ...(ii) The expression on the R.H.S. of (i) is divisible by 24. Thus Pk  1 is true whenever Pk  is true. Hence, by principle of mathematical induction, Pn is true for all n  N Illustration 4 Question: Prove the rule of exponents   n n n ab  a b by using principle of mathematical induction for every natural number. Solution: Let Pn be the given statement i.e,     n n n P n : ab  a b We note that Pn is true for n = 1 since   1 1 1 ab  a b Let Pk  be true, i.e.,   k k k ab  a b ...(i) We shall now prove that Pk  1 is true whenever Pk  is true. Now, we have ab ab ab k k  1 = a b  ab k k [by (i)] =    1 1 1 1 . . .    k k k k a a b b a b Therefore, Pk  1 is also true whenever Pk  is true. Hence, by principle of mathematical induction, Pn is true for all n  N . Illustration 5 Question: Using mathematical induction, show that   n N n n n           , 2 sin sin2 cos cos2 cos 4 ......cos 2 1 . Solution: Let             2 sin sin 2 : cos cos 2 cos 4 ......cos 2 1 n n n P n Step I: For n = 1 L.H.S. of (i) = cos  and R.H.S. of (i) =     cos 2 sin sin 2 Therefore, P(1) is true. Step II: Assume it is true for n = k, then             2 sin sin 2 : cos cos 2 cos 4 ......cos 2 1 k k k p k ...(i) Step III: For n = k+ 1
                  2 sin sin2 1 : cos cos2 cos4 ......cos 2 cos 2 1 1 1 k k k k P k L.H.S. =        k k cos cos 2 cos 4 ......cos 2 cos 2 1 =                 2 sin 2sin 2 cos 2 .cos 2 2 sin sin 2 k 1 k k k k k [using (i)] =      2 sin sin 2.2 k 1 k =       2 sin sin 2 1 1 k k = R.H.S. This shows that the Pk  1 is true if Pk  is true. Hence by the principle of mathematical induction, the result is true for all n  N . Illustration 6 Question: Prove by induction that the sum 3 5 3 3 2 Sn  n  n  n  is divisible by 3 for all n  N . Solution: Let Pn be the statement given by  : 3 5 3 3 2 P n Sn  n  n  n  is divisible by 3 Step I: We have, 1: 1 31 51 3 2 P S1    + 3 is divisible by 3 Since 1 31 51 3 12 3 2     , which is divisible by 3  P(1) is true. Step II: Let Pm be true. Then 3 5 3 3 2 Sm  m  m  m  is divisible by 3  3 5 3 3 2 Sm  m  m  m  = 3, for some   N ...(i) We now wish to show that Pm 1 is true. For this we have to show that  1 3 1 5 1 3 3 2 m   m   m   is divisible by 3 Now,  1 3 1 5 1 3  3 5 3 3 9 9 3 2 3 2 2 m   m   m    m  m  m   m  m  = 3 3 3 3 3[ 3 3] 2 2   m  m     m  m  [using (i)] = 3, where     m  3m  3 N 2  Pm 1 is true. Thus, Pm is true  Pm 1 is true Hence, by the principle of mathematical induction the statement is true for all n  N . Illustration 7 Question: Show by using principle of mathematical induction that   4 2 1 3 3 1.3 2.3 3.3 ...... .3 1 2 3        n n n n . Solution: Let     4 2 13 3 : 1.3 2.3 3.3 ...... .3 1 2 3        n n n P n n Step I: When n = 1, L.H.S. =1.3 = 3 and R.H.S. =     4 2.1 13 3 4 2 13 3 1 2      n n = 3 4 12  Hence P(1) is true. Let Pm be true    4 2 13 3 1.3 2.3 3.3 ...... .3 1 2 3        m m m m ....(i) To prove Pm 1 is true i.e.,     4 2 1 3 3 1.3 2.3 ...... .3 1.3 2 2 1           m m m m m m Adding   1 1 .3   m m to both sides of (i), we get   2 1 1.3 2.3 ...... .3 1 .3       m m m m =     1 1 1 .3 4 2 1 3 3       m m m m