PHYSICS-15-10- 11th (PQRS) SOLUTION

https://docs.google.com/document/d/1fznnCq_PLW8fjUCbguzODDwwiTamZe0R/edit?usp=sharing&ouid=109474854956598892099&rtpof=true&sd=true Mathematical Induction Principle of Mathematical Induction and its simple applications FIRST PRINCIPLE OF MATHEMATICAL INDUCTION The proposition P(n) involving a natural number n is assumed to be true for all n ∈ N, follows the following three steps: Step -I (Verification step) Actual verification of the proposition P(n) for the starting value of n = i. Step -II (Induction step) Assuming that if P(n) is true for n = k; k ≥ i, prove that it is also true for n = k + 1. Step -III (Generalization step) Combining the above two steps leads to the conclusion that P(n) is true for all integers n ∈ N. SECOND PRINCIPLE OF MATHEMATICAL INDUCTION (EXTENDED PRINCIPLE) Sometimes, the first principle of mathematical induction does not suffice. In such cases we use the extended principle as below: Step -I (Verification step) We verify that P(n) is true for i and (i + 1) both. Step -II (Induction step) Assume that P(n) is true for n = k and n = k + 1, k ≥ i, and prove that P(n) is true for n = (k + 2). Step -III (Generalization step) Combining the above two steps leads to the conclusion that P(n) is thus true n ∈ N. C H A P T E R CHAPTER INCLUDES : First principle of Mathematical Induction Second (extended) Principal of Mathematical Induction Application of Mathematical Induction Solved Examples APPLICATION OF MATHEMATICAL INDUCTION Identities Type Problems n5 + n3 + 7n Show by using induction that 5 3 15 is a natural number for n ∈ N. Solution : n5 + n3 + 7n Let P (n) : 5 3 15 is a natural number n5 + n3 + 7n = 15 + 13 + 7 = 1 + 1 + 7 = When n = 1, 5 3 15 5 3 15 5 3 15 1, which is a natural number Hence P (1) is true …(A) Let P (m) be true m5 + m3 + 7m ⇒ 5 3 15 is a natural number …(i) To prove P (m + 1) is true i.e., (m + 1)5 + (m + 1)3 + 5 3 Expanding (ii), we get 1 (m5 + 5m4 + 10m3 + 10m2 + 5m + 1) + 1 (m3 + 3m2 + 3m + 1) + 7 (m + 1) 5 = m5 + m3 + 7m + (m 4 + 2m3 + 3m2 + 3 2m) 15 + 1 + 1 + 7 5 3 15 5 3 15 = m5 + m3 + 7m + (m4 + 2m3 + 3m2 + 2m) + 1 5 3 15 ⎡ Θ m5 + m3 + 7m ⎤ = a natural number ⎢ 5 3 15 is a natural number from (i) ⎥ ⎥⎦ Hence P (m + 1) is true whenever P (m) is true …(B) From (A) and (B), it follows that P (n) is true for all natural numbers n. Divisibility Type Problems To prove that P(n) is divisible by x, following procedure is followed First we show that P(1) is divisible by x. Assuming that P(m) is divisible by x, it is proved that P(m + 1) is also divisible by x. For this either divide P(m + 1) with P(m) and show that remainder is divisible by x. or Split P(m + 1) = P(m) . I + R; I ∈ Z and show that R is divisible by x. Show that n3 + (n + 1)3 + (n + 2)3 is divisible by 9 for every natural number n. Solution : Let f (n) = n3 + (n + 1)3 + (n + 2)3 Let P (n) : f (n) i.e., n3 + (n + 1)3 + (n + 2)3 is divisible by 9. Now f (1) = 13 + 23 + 33 = 36, which is divisible by 9. ∴ P (1) is true. …(A) Let P (m) be true ⇒ f (m) = m3 + (m + 1)3 + (m + 2)3 is divisible by 9. ⇒ f (m) = m3 + (m + 1)3 + (m + 2)3 = 9k, where k is an integer …(i) To prove P (m + 1) is true i.e., f (m + 1) is divisible by 9. Now, f (m + 1) = (m + 1)3 + (m + 2)3 + (m + 3)3 = (m + 1)3 + (m + 2)3 + m3 + 9m2 + 27m + 27 = [m3 + (m + 1)3 + (m + 2)3] + 9m2 + 27m + 27 = 9k + 9 (m2 + 3m + 3), which is divisible by 9. Hence P (m + 1) is true whenever P (m) is true. …(B) From (A) and (B), it follows that P (n) is true for all natural numbers n. Inequalities Type Problems If a and b are positive, use mathematical induction to prove that ⎛ a + b ⎞n an + bn ⎜ ⎟ ≤ ⎝ 2 ⎠ Solution : 2 ∀ n ∈ N . ⎛ a + b ⎞n an + bn P (n) : ⎜ ⎟ ≤ ⎝ 2 ⎠ 2 Step I. For n = 1 P(i ) : a + b ≤ a + b 2 2 which is true for n = 1, Step II. Assume it is true for n = k, then ⎛ a + b ⎞k ak + bk P(k ) : ⎜ ⎟ ≤ ⎝ 2 ⎠ 2 Step III. For n = k + 1, ⎛ a + b ⎞k+1 ⎛ a + b ⎞ ⎛ a + b ⎞k ⎛ a + b ⎞ ⎛ ak + bk ⎞ ⎜ ⎟ = ⎜ ⎟ ⎜ ⎟ ≤ ⎜  ⎟ ⎜ ⎟ (By assumption step) ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎝ ⎟ ⎛ a + b ⎞k +1 1 ⇒ ⎜ ⎟ ⎝ ⎠ ≤ [(ak+1 + bk+1) + ak b + abk ] 4 ⎛ a + b ⎞k+1 1 ⇒ ⎜ ⎟ ⎝ ⎠ ≤ [2 (ak +1 + bk+1) – ak +1 – bk+1 + ak b + abk ] 4 ⎛ a + b ⎞k +1 1 ⇒ ⎜ ⎟ ⎝ ⎠ ≤ [2 (ak +1 + bk +1) – (a − b) (ak − bk )] 4 ⎛ k +1 k +1 ⎞ ≤ ⎜ ⎟ − (a − b) (ak − bk ) ⎜ ⎟ ⎝ ⎠ ⎛ a + b ⎞k +1 ak +1 + bk +1 ∴ ⎜ ⎟ ≤ ⎝ ⎠ 2 Hence the statement is true for n = k + 1 and by the principle of induction it is true for all n ∈ N. Problems Based on Extended Principle of Mathematical Induction If the given problem cannot be solved by direct use of principle mathematical induction, try to use the extended principle of mathematical induction. If a1 = 1, a2 = 5 and an+2 = 5an+1 – 6an, n ≥ 1. Show by using mathematical induction that an = 3n – 2n. Solution : Let P (n) : an = 3n – 2n When n = 1, L.H.S. = a1 = 1 (given) and R.H.S. = 31 – 21 = 3 – 2 = 1. ∴ L.H.S. = R.H.S. When n = 2, L.H.S. = a2 = 5 (given) and R.H.S. = 32 – 22 = 5 Thus L.H.S = R.H.S. Hence P (1) and P (2) are true …(A) Let P (m) and P (m + 1) be true ⎧⎪am = 3m − 2m ⇒ ⎨ ⎩am+1 m +1 − 2m +1 …(ii) To prove P (m + 2) is true i.e., am+2 = 3m+2 – 2m+2 Given, an+2 = 5an+1 – 6an ∴ am+2 = 5am+1 – 6am = 5 (3m+1 – 2m+1) – 6 (3m – 2m) = (5.3m+1 – 6.3m) – (5.2m+1 – 6.2m) = 3m (15 – 6) – 2m (10 – 6) = 3m.9 – 2m.4 = 3m+2 – 2m+2. Hence P (m + 2) is true whenever P (m) and P (m + 1) are true. …(B) From (A) and (B), by the principle of mathematical induction it follows that P (n) is true for all natural numbers n. SOLVED EXAMPLES Example 1 : Use the principle of mathematical induction to prove that for all n ∈ N. = 2cos ⎛ π ⎞ ⎜ 2n +1 ⎟ ⎝ ⎠ when the L.H.S. contains n radical signs. Solution : = 2cos ⎛ π ⎞ Let P (n) = ⎜ 2n +1 ⎟ …(i) Step I. For n = 1, L.H.S. of (i) = and R.H.S. of (i) ⎝ ⎠ = 2 cos ⎛ π ⎞ 2 ⎝ ⎠ = 2 cos ⎛ π ⎞ ⎜ ⎟ ⎝ 4 ⎠ Therefore, P (1) is true. Step II. Assume it is true for n = k, P (k ) = = 2cos ⎛ π ⎞ ⎜ 2k +1 ⎟ (k radical sign) ⎝ ⎠ Step III. For n = k + 1 P (k + 1) = (k + 1) radical sign = = (By assumption step) = = = = 2 cos ⎛ π ⎞ ⎜ 2k +2 ⎟ ⎝ ⎠ This shows that the result is true for n = k + 1. Hence by the principle of mathematical induction, the result is true for all n ∈ N. Prove that 1 + + +… + ≥ , n ∈ N. Solution : For n = 1, L.H.S. = 1, R.H.S. = 1 so L.H.S. = R.H.S. …(i) Assume the result for n = k, i.e., 1+ + +… + 1 ≥ …(ii) For n = k + 1, L.H.S. = 1+ + …+ 1 + ≥ + [Using (ii)] + [Note] = + − = i.e., the result is true for n = k + 1 Hence by induction, the result is true n ∈ N. ❑ ❑ ❑