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Principle of Mathematical Induction

Long 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 $straight n space element of space straight N.$

Let P(n): $41 to the power of straight n space minus space 14 to the power of straight n$ is a multiple of 27.

I.       For n = 1,

        P(1): $41 to the power of 1 space minus space 14 to the power of 1$ is a multiple of 27

$rightwards double arrow$    41 - 14 is a multiple of 27 $rightwards double arrow$ 27 is a multiple of 27

         which is true.

∴ P(n) is true for n = 1

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

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III. For n = m + 1,

 P (m + 1) : $41 to the power of straight m plus 1 end exponent space minus space 14 to the power of straight m plus 1 end exponent$ is a multiple of 27

 But, $41 to the power of straight m plus 1 end exponent minus space 1 space 4 to the power of straight m plus 1 end exponent space equals space 41 to the power of straight m plus 1 end exponent space minus space 41 to the power of straight m. end exponent 14 space plus space 41 to the power of straight m.14 space minus space 14 to the power of straight m plus 1 end exponent$

 = $41 to the power of straight m left parenthesis 41 minus 14 right parenthesis space plus 14 space left parenthesis 41 to the power of straight m minus 14 to the power of straight m right parenthesis space equals space 41 to the power of straight m space cross times space 27 space plus space 14 space left parenthesis 27 straight k right parenthesis$ [By (i)]

 $equals 27 space left square bracket 41 to the power of straight m space plus space 14 straight k right square bracket space equals space 27 straight k apostrophe$ where $straight k apostrophe space equals space 41 to the power of straight m space plus space 14 space straight k space element of space straight Z$

$rightwards double arrow space space space space space space space space space space 41 to the power of straight m plus 1 end exponent space minus space 14 to the power of straight m plus 1 end exponent$ is a multiple of 27

∴ P(m + 1) is true.

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

Hence, by induction, P(n) is true 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.