Tiêu đề XI - maths - chapter 4 - MATHEMATICAL INDUCTION (68-83).pdf ✅

MATHEMATICAL INDUCTION 68 NARAYANAGROUP JEE-MAIN-JR-MATHS VOL-I  Principle of Finite Mathematical Induction: For n N  , let P(n) be a statement in terms of n. If P(1) is true and P(k) is true  P(k + 1) is true, then P(n) is true, for all n N  .  Principle of Complete Mathematical Induction: For n N  , let P(n) be a statement in terms of n. If P(1), P(2), P(3),.... P(k-1) are true  P(k) is true, then P(n) is true, for all n N  .  a a d a d , , 2 , ...       form an A.P. then (i) nth term t a n d n     1 , Where a is the first term and d is the common difference. (ii) Sum of n terms 2 1   2 n n S a n d          2 n   a l Where a = first term, l = last term  2 , a a r a r , , ... form a G.P then (i) nth term 1 . n nt a r   , Where a = first term r = common ratio (ii) Sum of n terms  1; 1 n n r S a r    (iii) In an infinite G.P, Sum of Infinite terms is 1 a S r    where r 1      2 a a d r a d r      2 ...   1 ...... 1 n a n d r         form A.G.P. then (i) nth term   1 1 n nt a n d r         (ii) Sum of n terms         1 2 1 1 1 1 1 n n n a dr r a n d r S r r r               (iii) Sum of Infinite terms   2 1 1 1 a dr S where r r r       W.E-1: Sum to infinite terms of 2 3 1 3 5 7 .....       x x x (If x 1)is__ [Eam-1998] Sol:   2 1 1 a dr s r r      where a 1, d  2 ,r x      2 2 1 2 1 1 1 1 x x x x x        Similarly If x 1,   2 3 2 1 1 2 3 4 ... 1 s x x x x           SOME IMPORTANT POINTS i) Sum of first n natural numbers i.e. n n 1   n 1 2 3 ......n 2        , n N ii) Sum of the squares of first n natural numbers is    2 2 2 2 2 1 2 1 1 2 3 ... , 6 n n n n n n N            iii) Sum of the cubes of first n natural numbers is   2 2 3 3 3 3 3 1 1 2 3 ... 4 n n n n           2    n n N , iv) 4 4 4 4 1 2 3 ......      n 4 n      4 2 1 1 3 1 5 6 n n n n n n       v) Sum of the first ‘n’ odd +ve integers = 1 + 3 + 5 + ... + (2n-1) 2  n vi) Sum of the first ‘n’ even +ve integers = 2 + 4 + 6 + ............ + 2n   n n 1 MATHEMATICAL INDUCTION SYNOPSIS


MATHEMATICAL INDUCTION NARAYANAGROUP 71 JEE-MAIN-JR-MATHS VOL-I 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 conclude 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 1) 6n  4 n 2) 3n  2 n 3) 4n  6 n 4) 2(2 3) 1 n  9. 2+7+14+......+ (n2 + 2n - 1) = 1) 6 (2 9 1) 2 n n  n  2) 6 2 9 1 2 n  n  3) 12 2 9 1 2 n  n  4) 24 2 9 1 2 n  n  10. 1+3+6+10+ ........+         2 1 2 n 1 n n n 1)    3 n n 1 n  2 2)    6 n 1 n  2 3)    6 n n 1 n  2 4)    3 2 1 2 n  n  11. 3.6+6.9+9.12+.......+ 3n (3n+3) = 1) 3 n(n 1)(n  2) 2) 3n (n+1) (n+2) 3) 3 (n1)(n2)(n3) 4) 4 (n 1)(n  2)(n  4) LEVEL - I (C.W)