Prove the following by using the principle of mathematical induc

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 Multiple Choice QuestionsLong Answer Type

11.

Prove the following by using the principle of mathematical induction for all straight n element of space straight N.

a + (a + d) + (a + 2d) + ...........+ [a + (n - 1)d] = straight n over 2 left square bracket 2 straight a plus left parenthesis straight n minus 1 right parenthesis straight d right square bracket



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12.

Prove the following by using the principle of mathematical induction for all straight n space element of space straight N:

open parentheses 1 plus 1 over 1 close parentheses open parentheses 1 plus 1 half close parentheses open parentheses 1 plus 1 third close parentheses space........ open parentheses 1 plus 1 over straight n close parentheses space equals space left parenthesis straight n space plus space 1 right parenthesis


Let P(n): open parentheses 1 plus 1 over 1 close parentheses open parentheses 1 plus 1 half close parentheses open parentheses 1 plus 1 third close parentheses....... open parentheses 1 plus 1 over straight n close parentheses space equals space left parenthesis straight n plus 1 right parenthesis

I.       
For  n = 1,

        P(1) : open parentheses 1 plus 1 over 1 close parentheses space equals space 1 space plus space 1 space rightwards double arrow space 1 plus space 1 space equals space 1 plus space 1 space rightwards double arrow space 2 space equals space 2

∴       P(1) is true

II.      Suppose that the statement P (n) is true for n = m,  straight m space element of space straight N

∴      P(m) : open parentheses 1 plus 1 over 1 close parentheses open parentheses 1 plus 1 half close parentheses open parentheses 1 plus 1 third close parentheses....... open parentheses 1 plus 1 over straight m close parentheses space equals space straight m plus space 1                 ...(i)

III.     For n = m + 1,

        straight P space left parenthesis straight m space plus space 1 right parenthesis colon space open parentheses 1 plus 1 over 1 close parentheses open parentheses 1 plus 1 half close parentheses open parentheses 1 plus 1 third close parentheses........ space open parentheses 1 plus fraction numerator 1 over denominator straight m plus 1 end fraction close parentheses equals space straight m space plus space 1 space plus space 1

or      space space open square brackets open parentheses 1 plus 1 over 1 close parentheses open parentheses 1 plus 1 half close parentheses open parentheses 1 plus 1 third close parentheses...... open parentheses 1 plus 1 over straight m close parentheses close square brackets space space open parentheses 1 plus fraction numerator 1 over denominator straight m plus 1 end fraction close parentheses space equals space straight m space plus space 2

       From (i), open parentheses 1 plus 1 over 1 close parentheses open parentheses 1 plus 1 half close parentheses open parentheses 1 plus 1 third close parentheses space....... open parentheses 1 plus 1 over straight m close parentheses space equals space straight m space plus space 1

∴   straight P space left parenthesis straight m space plus space 1 right parenthesis colon space left parenthesis straight m space plus space 1 right parenthesis space open parentheses 1 plus fraction numerator 1 over denominator straight m plus 1 end fraction close parentheses space space equals space straight m space plus space 2 space space rightwards double arrow space left parenthesis straight m space plus space 1 right parenthesis space open parentheses fraction numerator straight m plus 1 plus 1 over denominator straight m plus 1 end fraction close parentheses space equals space straight m space plus space 2

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              which is true
∴             P (m + 1) is true

∴             P(m) is truerightwards double arrowP (m + 1) is true

Hence, by the principle of mathematical induction,  P(n) is true for all straight n element of space straight N.

         



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13.

Prove the following by using the principle of mathematical induction for all straight n space element of space straight N.

space space 1 plus 2 plus 3 plus........ plus straight n less than 1 over 8 left parenthesis 2 straight n plus 1 right parenthesis squared

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14.

Prove the following by using the principle of mathematical induction for all straight n space element of space straight N.

space space fraction numerator 1 over denominator straight n plus 1 end fraction space plus space fraction numerator 1 over denominator straight n plus 2 end fraction space plus space.......... space plus space fraction numerator 1 over denominator 2 straight n end fraction greater than 13 over 24


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15.

Prove the following by using the principle of mathematical induction for all space space straight n element of straight N.

n (n + 1) (n + 5) is a multiple of 3.

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16.

Prove the following by using the principle of mathematical induction for all straight n space element of space straight N.

41 to the power of straight n space minus space 14 to the power of straight n is a multiple of 27 for all straight n space element of space straight N.



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17.

Prove the following by using the principle of mathematical induction for all straight n space element of space straight N.

10 to the power of 2 straight n minus 1 end exponent space plus space 1 is divisible by 11.

144 Views

18.

Prove the following by using the principle of mathematical induction for all straight n space element of space straight N.

3 to the power of 2 straight n plus 2 end exponent minus 8 straight n minus 9 is divisible by 8.

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19.

Prove by mathematical induction that sum of cubes of three consecutive natural numbers is divisible by 9.

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20.

Show by mathematical induction that a2n – b2n is divisible by a + b.

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