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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] = $straight n over 2 left square bracket 2 straight a plus left parenthesis straight n minus 1 right parenthesis straight d right square bracket$

265 Views

12.

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

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

137 Views

13.

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

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

124 Views

14.

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

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

265 Views

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.

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

I. For n = 1,

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$rightwards double arrow$ 1 (2) (6) is a multiple of 3 $rightwards double arrow$ 12 is a multiple of 3.

 which is true.

∴ P(n) is true for n = 1.

II. Suppose P (n) is true for n = m

$rightwards double arrow$ P(m) : m(m + 1) (m + 5) is a multiple of 3 $rightwards double arrow$ m (m + 1) (m + 5) = 3k

$rightwards double arrow$ $straight m cubed plus 6 straight m squared plus 5 straight m space equals space 3 straight k$

$rightwards double arrow$ $straight m cubed space equals space 3 straight k space minus space 6 straight m squared minus 5 straight m$ ...(i)

III. For n = m + 1,

 P(m + 1): (m + 1) (m + 1 + 1) (m + 1 + 5) is a multiple of 3.

 (m + 1) (m + 2) (m + 6) is a multiple of 3.

 Now, (m + 1) (m + 2) (m + 6) = (m + 1)( $straight m squared plus 8 straight m plus 12$ ) =

 $straight m cubed plus 9 straight m squared plus 20 straight m plus 12$
 = $3 straight k minus 6 straight m squared minus 5 straight m space plus space 9 straight m squared space plus space 20 straight m space plus space 12$ [BY (i)]

 = $3 straight k space plus space 3 straight m squared space plus space 15 straight m space plus space 12 space equals space 3 left square bracket straight k plus straight m squared plus 5 straight m plus 4 right square bracket equals 3 straight k apostrophe$

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$rightwards double arrow$ (m + 1) (m + 2) (m + 6) is a multiple of 3.

∴ P(m + 1) is true.

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

Hence, P(n) is true for all $straight n element of space straight N.$

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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 $straight n space element of space straight N.$

171 Views

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.