Tiêu đề XI - maths - chapter 4 - MATHEMATICAL INDUCTION(11.03.2015)(83-100).pdf ✅

86 MATHEMATICAL INDUCTION JEE MAINS - VOL - I LEVEL - I (C.W) 14. The greatest +ve integer which divides n n n r    1 2 .......    , for all n N  is 1) r 1 ! 2) r! 3) r 4) r-1 C.U.Q-KEY 01) 2 02) 4 03) 3 04) 2 05) 2 06) 1 07) 4 08) 1 09) 3 10) 2 11) 3 12) 2 13) 1 14) 2 C.U.Q-HINTS 1. By the definition of P.OF.M.I 2. By the definition of P.OF.M.I 3. When n LH S RHS     1, . . . 1.1!, . . 2! 1! when n LH S RH S      2, . . . 1.1! 2.2!, . . 3! 1 Hence P n  is true  n N 4. By Verification 5. By verification 6. The product of 2 consecutive numbers is always even. 7. It is obvious. 8. Put n = 1,2 and verify the options. 9. 1 a   7 7 ; Let 7 ma  .Then a a m m 1  7 2 1 7 7 7 14 m m a a       1 14 7 ma    ; So 7, n a n   10. By verification 11. Put n = 1, n = 2 and verify the options. 12.   n n P n a b n N     put n 1,  P a b 1   which is divisible by a + b put n  2 ,    2 2 P a b 2   not divisible by a b  , put n = 3       3 3 2 2 P a b a b a ab b 3       which is divisible by a b  . With the help of induction we conclude that P n  will be divisible by a b  if n is odd. 13. put n  4 and P  2 . 14. Product of r successive integers is divisible byr! Principle of Mathematical Induction: 1. A student was asked to prove a statement by induction. He proved (i) P(5) is true and (ii) truth of P(n)=> truth of P(n+1), n N  . On the basis of this, he could con- clude that P(n) is true 1) for no n N  2) for all n n N   5, 3) for all n N  4) for all n n N   1, 2. If P n  be the statement n n   1 1  is an integer, then which of the following is even 1) P2 2) P3 3) P4 4) None of the above 3. n > 1, n even  digit in the units place of 2 2 1 n  1) 5 2) 7 3) 6 4) 1 4. log .log   n x n x  is true for n . 1)  n N 2)  n Z 3) n is positive odd integer 4) n is positive even integer Inequalities: 5.   2 ! n n n  is true for 1)  n N 2)    n n N 1, 3)    n n N 2, 4)  n Z 6. Let   2 1 1 1 1 :1 ..... 2 4 9 P n n n       is true for 1)  n N 2) n 1 3) n n N    1, 4) n  2 Summation of Series: 7. 3 3 3 3 3 1 2 3 4 ..... 9       1) 425 2) -425 3) 475 4) -475 8. 1 1 1 ..... 2.5 5.8 8.11     n terms