Prove by mathematical induction that sum of cubes of three conse

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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

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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.

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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.


Let n, n+1, n+2 be three consecutive natural numbers.

Let P(n): straight n cubed space plus space left parenthesis straight n plus 1 right parenthesis cubed space plus space left parenthesis straight n plus 2 right parenthesis cubed is divisible by 9.

I.       For n = 1,

       straight P left parenthesis 1 right parenthesis space colon space 1 cubed space plus space 2 cubed space plus space 3 cubed is divisible by 9

rightwards double arrow  1 + 8 + 27 is divisible by 9 rightwards double arrow36 is divisible by 9

      which is true

∴     the statement is true for n = 1.

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

rightwards double arrow   P(m) : straight m cubed space plus space left parenthesis straight m space plus space 1 right parenthesis cubed space plus space left parenthesis straight m space plus space 2 right parenthesis cubed is divisible by 9.

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#6 {main}</pre>                           ...(i)

III.   For n = m + 1,                           

       straight P left parenthesis straight m plus 1 right parenthesis space colon space left parenthesis straight m plus 1 right parenthesis cubed space plus space left parenthesis straight m plus 2 right parenthesis cubed space plus space left parenthesis straight m plus 3 right parenthesis cubed is divisible by 9.

       Now, from (i),
     
       space space straight m cubed space plus space left parenthesis straight m plus 1 right parenthesis cubed space plus space left parenthesis straight m plus 2 right parenthesis cubed space equals space 9 straight k      

rightwards double arrow     left parenthesis straight m plus 1 right parenthesis cubed plus left parenthesis straight m plus 2 right parenthesis cubed space equals space 9 straight k space minus space straight m cubed space space rightwards double arrow space left parenthesis straight m plus 1 right parenthesis cubed space plus space left parenthesis straight m plus 2 right parenthesis cubed space plus space left parenthesis straight m space plus space 3 right parenthesis cubed space equals space 9 straight k space minus space straight m cubed space plus space left parenthesis straight m plus 3 right parenthesis cubed

         negative 9 straight k minus straight m cubed plus straight m cubed plus 27 space plus space 3 straight m squared left parenthesis 3 right parenthesis space plus space 3 space left parenthesis 9 right parenthesis space straight m space equals space 9 straight k plus 27 plus 9 straight m squared plus 27 straight m space equals space 9 left parenthesis straight k plus 3 plus straight m squared plus 3 straight m right parenthesis

         where straight k space element of space straight Z comma space space space straight m space element of space straight N

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#6 {main}</pre> is divisible by 9

rightwards double arrow       P (m + 1) is true

∴         P (m) is true rightwards double arrow P (m + 1) is true.

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






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20.

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

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