Prove the following by using the principle of mathematical induc

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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




Let P(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

I.     For n = 1,

      straight P left parenthesis 1 right parenthesis colon space straight a space equals space 1 half left square bracket 2 straight a space plus space left parenthesis 1 minus 1 right parenthesis straight d right square bracket

rightwards double arrow   straight a equals 1 half left square bracket 2 straight a space plus 0 right square bracket space rightwards double arrow space straight a space equals space 1 half left parenthesis 2 straight a right parenthesis space rightwards double arrow space straight a space equals space straight a

∴      P(1) is true

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

∴       straight P left parenthesis straight m right parenthesis colon space straight a space plus space left parenthesis straight a space plus space straight d right parenthesis space plus space left parenthesis straight a space plus space 2 straight d right parenthesis space plus space left curly bracket straight a space plus space left parenthesis straight m space minus space 1 right parenthesis straight d right curly bracket space equals space straight m over 2 left square bracket 2 straight a space plus space left parenthesis straight m minus 1 right parenthesis straight d right square bracket .... (i)

III.   For n = m + 1,

       space space straight P left parenthesis straight m space plus space 1 right parenthesis space colon space straight a space plus space left parenthesis straight a space plus space straight d right parenthesis space plus space left parenthesis straight a space plus space 2 straight d right parenthesis space plus space....... space plus space left parenthesis straight a space plus space md right parenthesis space equals space fraction numerator straight m plus 1 over denominator 2 end fraction left square bracket 2 straight a space plus space left parenthesis straight m plus 1 minus 1 right parenthesis space straight d right square bracket

or      straight a plus left parenthesis straight a plus straight d right parenthesis plus left parenthesis straight a plus 2 straight d right parenthesis space plus space............ plus space left curly bracket straight a space plus space left parenthesis straight m minus 1 right parenthesis straight d right curly bracket space plus space left parenthesis straight a space plus space md right parenthesis space equals space fraction numerator straight m plus 1 over denominator 2 end fraction left square bracket 2 straight a space plus space md right square bracket

       From (i),

       straight a plus left parenthesis straight a plus straight d right parenthesis plus left parenthesis straight a plus 2 straight d right parenthesis plus.......... plus left curly bracket straight a plus left parenthesis straight m minus 1 right parenthesis straight d right curly bracket space equals space straight m over 2 left square bracket 2 straight a space plus space left parenthesis straight m space minus space 1 right parenthesis straight d right square bracket

∴      straight P left parenthesis straight m plus 1 right parenthesis space colon space straight m over 2 left square bracket 2 straight a space plus space left parenthesis straight m minus 1 right parenthesis space straight d right square bracket space plus space left parenthesis straight a space plus space md right parenthesis space equals space fraction numerator straight m plus 1 over denominator 2 end fraction left square bracket 2 straight a plus space md right square bracket

rightwards double arrow space space space space 1 half left square bracket 2 ma space plus space straight m squared straight d space minus space md space plus space 2 straight a space plus space 2 md right square bracket space equals space fraction numerator straight m space plus space 1 over denominator 2 end fraction left square bracket 2 straight a space plus space md right square bracket

rightwards double arrow space space space 1 half left square bracket 2 ma space plus space 2 straight a space plus space straight m squared straight d space plus space md right square bracket space equals space fraction numerator straight m plus space 1 space over denominator 2 end fraction left parenthesis 2 straight a space plus space md right parenthesis

space space rightwards double arrow space space 1 half left square bracket 2 straight a space left parenthesis straight m space plus space 1 right parenthesis space plus space md space left parenthesis space straight m plus space 1 right parenthesis right square bracket space equals space fraction numerator straight m plus 1 over denominator 2 end fraction left parenthesis 2 straight a space plus space md right parenthesis

rightwards double arrow space space space space space space fraction numerator straight m plus 1 over denominator 2 end fraction left square bracket 2 straight a space plus space md right square bracket space equals space fraction numerator straight m plus 1 over denominator 2 end fraction left square bracket 2 straight a space plus space md right square bracket

           which is true

∴          P (m + 1) is true

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

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20.

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

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