Principle of Mathematical Induction

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Question 1

If A = open square brackets table row 1 0 row 1 1 end table close square brackets and I = open square brackets table row 1 0 row 0 1 end table close square brackets , then which one of the following holds for all n ≥ 1, by the principle of mathematical induction

  • An = nA – (n – 1)I

  • An = 2n-1A – (n – 1)I

  • An = nA + (n – 1)I

  • An = 2n-1A + (n – 1)I

Solution

A.

An = nA – (n – 1)I

By the principle of mathematical induction (1) is true.

Question 2

If the number of terms in the expansion of open parentheses 1 minus 2 over straight x space plus 4 over straight x squared close parentheses to the power of straight n comma straight x not equal to 0 comma is 28, then the sum of the coefficients of all the terms in this expansion is

  • 64

  • 2187

  • 243

  • 729

Solution

D.

729

Clearly, number of terms in the expansion of 

open parentheses 1 minus 2 over straight x plus 4 over straight x squared close parentheses space is space fraction numerator left parenthesis straight n plus 2 right parenthesis left parenthesis straight n plus 1 right parenthesis over denominator 2 end fraction space or space straight C presuperscript straight n plus 2 end presuperscript subscript 2
left square bracket assuming space 1 over straight x space and space 1 over straight x squared space distinct right square bracket
therefore comma space fraction numerator left parenthesis straight n plus 2 right parenthesis left parenthesis straight n plus 1 right parenthesis over denominator 2 end fraction space equals space 28
rightwards double arrow space left parenthesis straight n plus 2 right parenthesis left parenthesis straight n plus 1 right parenthesis space equals space 56 space
space equals space left parenthesis 6 plus 1 right parenthesis left parenthesis 6 plus 2 right parenthesis space rightwards double arrow straight n space equals space 6
Hence comma space sum space of space coefficients equals space left parenthesis 1 minus 2 plus 4 right parenthesis to the power of 6 space equals space 3 to the power of 6 space equals space 729

Question 3

Let S(K) = 1 +3+5+..... (2K-1) = 3+K2. Then which of the following is true?

  • S(1) is correct

  • Principle of mathematical induction can be used to prove the formula

  • S(K) ≠S(K+1)

  • S(K)⇒ S(K+1)

Solution

D.

S(K)⇒ S(K+1)

S(K) = 1 + 3 + 5 + ...... + (2K - 1) = 3 + K2
Put K = 1 in both sides
∴ L.H.S = 1 and R.H.S. = 3 + 1 = 4 ⇒ L.H.S. ≠ R.H.S.
Put (K + 1) on both sides in the place of K L.H.S. = 1 + 3 + 5 + .... + (2K - 1) + (2K + 1)
R.H.S. = 3 + (K + 1)2 = 3 + K2 + 2K + 1
Let L.H.S. = R.H.S.
1 + 3 + 5 + ....... + (2K - 1) + (2K + 1) = 3 + K2 + 2K + 1
⇒ 1 + 3 + 5 + ........ + (2K - 1) = 3 + K2 If S(K) is true, then S(K + 1) is also true. Hence, S(K) ⇒ S(K + 1)

Question 4

Maximum sum of coefficient in the expansion of (1 – x sinθ + x2 )n is

  • 1

  • 2n

  • 3n

  • 0

Solution

C.

3n

Sum of coefficients in (1 – x sinθ + x2 )n is (1 – sinθ + 1)n
(putting x = 1)
This sum is greatest when sinθ = –1, then maximum sum is 3n .

Question 5

Statement − 1: For every natural number n ≥ 2 fraction numerator 1 over denominator square root of 1 end fraction space plus space fraction numerator 1 over denominator square root of 2 end fraction space plus space..... space plus space fraction numerator 1 over denominator square root of straight n end fraction space greater than space square root of straight n

Statement −2: For every natural number n ≥ 2,straight n greater or equal than 2 comma space square root of straight n left parenthesis straight n plus 1 right parenthesis space end root space less than space straight n plus 1

  • Statement −1 is false, Statement −2 is true

  • Statement −1 is true, Statement −2 is true, Statement −2 is a correct explanation for Statement −1

  • Statement −1 is true, Statement −2 is true; Statement −2 is not a correct explanation for Statement −1.

  • Statement − 1 is true, Statement − 2 is false. 

Solution

C.

Statement −1 is true, Statement −2 is true; Statement −2 is not a correct explanation for Statement −1.

straight P space left parenthesis straight n right parenthesis space equals space fraction numerator 1 over denominator square root of 1 end fraction space plus space fraction numerator 1 over denominator square root of 2 end fraction space plus space..... space plus space fraction numerator 1 over denominator square root of straight n end fraction
straight P space left parenthesis 2 right parenthesis space equals space fraction numerator 1 over denominator square root of 1 end fraction space plus space fraction numerator 1 over denominator square root of 2 end fraction space greater than space square root of 2
Let space us space assume space that space straight P space left parenthesis straight k right parenthesis space equals space fraction numerator 1 over denominator square root of 1 end fraction space plus space fraction numerator 1 over denominator square root of 2 end fraction space plus space...... fraction numerator 1 over denominator square root of straight k end fraction space plus fraction numerator 1 over denominator square root of straight k plus 1 end root end fraction space greater than square root of straight k plus 1 end root
has space to space be space true.
straight L. straight H. straight S greater than thin space square root of straight k space plus space fraction numerator 1 over denominator square root of straight k plus 1 end root end fraction space equals space fraction numerator square root of straight k space left parenthesis straight k plus 1 right parenthesis end root space plus 1 over denominator square root of straight k plus 1 end root end fraction
since space square root of straight k left parenthesis straight k plus 1 right parenthesis end root space greater than straight k space space left parenthesis for all space straight k greater or equal than 0 right parenthesis
therefore space fraction numerator square root of straight k space left parenthesis straight k plus 1 right parenthesis end root plus 1 over denominator square root of straight k plus 1 end root end fraction space greater than space fraction numerator straight k plus 1 over denominator square root of straight k plus 1 end root end fraction space equals space square root of straight k plus 1 end root
Let space straight p left parenthesis straight n right parenthesis space space equals space square root of straight n space left parenthesis straight n plus 1 right parenthesis end root space less than space straight n plus 1
State space minus 1 space is space correct.
straight P space left parenthesis 2 right parenthesis space space equals space square root of 2 space straight x space 3 end root space less than space 3
If space straight P space left parenthesis straight k right parenthesis space space equals space square root of straight k left parenthesis straight k plus 1 right parenthesis end root space less than space left parenthesis straight k plus 1 right parenthesis space is space true
Now space space straight P space left parenthesis straight k plus 1 right parenthesis space equals space square root of left parenthesis straight k plus 1 right parenthesis left parenthesis straight k plus 2 right parenthesis end root space less than space straight k plus 2 space has space to space be space true
square root of left parenthesis straight k plus 1 right parenthesis left parenthesis straight k plus 2 right parenthesis end root space thin space left parenthesis straight k plus 2 right parenthesis
Hence Statement −2 is not a correct explanation of Statement −1.