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

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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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#6 {main}</pre> is divisible by 8.

I.      For n = 1,

      P(1) : 3 to the power of 2 left parenthesis 1 right parenthesis plus 2 end exponent space minus space 8 space left parenthesis 1 right parenthesis space minus space 9 is divisible by 8

rightwards double arrow  3 to the power of 4 minus 8 minus 9 is divisible by 8 rightwards double arrow 81 - 17 is divisible by 8 rightwards double arrow 64 is divisible by 8

      which is true

∴     P(n) is true for n = 1

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#6 {main}</pre>

∴     straight P left parenthesis straight m right parenthesis space colon space 3 to the power of 2 straight m plus 2 end exponent space minus space 8 straight m space minus space 9 is divisible by 8

rightwards double arrow   space space space space 3 to the power of 2 straight m plus 2 end exponent space minus space 8 straight m space minus space 9 space equals space 8 straight k comma space straight k space element of space straight Z space rightwards double arrow space 3 to the power of 2 straight m plus 2 end exponent space equals space 8 straight k space plus space 8 straight m space plus space 9   ...(i)

III.     For n = m + 1,

        space space space straight P left parenthesis straight m plus 1 right parenthesis colon space 3 to the power of 2 left parenthesis straight m plus 1 right parenthesis plus 2 end exponent space minus space 8 left parenthesis straight m space plus space 1 right parenthesis space minus space 9 is divisible by 8

Now,    3 to the power of 2 left parenthesis straight m plus 1 right parenthesis plus 2 end exponent space minus space 8 left parenthesis straight m plus 1 right parenthesis minus 9 space equals space 3 to the power of 2 straight m plus 2 end exponent space. space 3 squared minus 8 straight m minus 8 minus 9

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#6 {main}</pre>                                      [By (i)]

= 72k + 72m + 81 - 8m - 17 = 72k + 64m + 64 = 8(9k + 8m + 8), <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
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#6 {main}</pre>

= 8k'  where k' = 9k + 8m + 8 space element of space straight Z

∴     3 to the power of 2 left parenthesis straight m plus 1 right parenthesis plus 2 end exponent space minus space 8 left parenthesis straight m space plus space 1 right parenthesis space minus space 9 is divisible by 8.

rightwards double arrow  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 space element of space straight N.

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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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