Prove the following by using the principle of mathematical induc

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 Multiple Choice QuestionsLong Answer Type

1.

Use principle of mathematical induction to prove that:

1 space plus space 2 space plus space 3 space plus space... space plus space straight n space equals space fraction numerator straight n left parenthesis straight n space plus space 1 right parenthesis over denominator 2 end fraction

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2.

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

1 cubed plus 2 cubed plus 3 cubed plus space... space plus space straight n cubed space equals space open square brackets fraction numerator straight n left parenthesis straight n plus 1 right parenthesis over denominator 2 end fraction close square brackets squared

596 Views

3.

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

1 plus 3 plus 3 squared plus....... space plus 3 to the power of straight n minus 1 end exponent space equals space fraction numerator 3 to the power of straight n minus 1 over denominator 2 end fraction

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4.

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

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#6 {main}</pre>

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5.

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

fraction numerator 1 over denominator 3.5 end fraction space plus space fraction numerator 1 over denominator 5.7 end fraction space plus space fraction numerator 1 over denominator 7.9 space end fraction space plus space......... space plus space fraction numerator 1 over denominator left parenthesis 2 straight n plus 1 right parenthesis left parenthesis 2 straight n plus 3 right parenthesis end fraction space equals space fraction numerator straight n over denominator 3 space left parenthesis 2 straight n space plus space 3 right parenthesis end fraction

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6.

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

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#6 {main}</pre>




226 Views

7.

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

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221 Views

8.

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

1.3 space plus space 2.3 squared space plus space 3.3 cubed space plus space........ space plus space straight n.3 to the power of straight n space equals space fraction numerator left parenthesis 2 straight n minus 1 right parenthesis space 3 to the power of straight n plus 1 end exponent space plus space 3 over denominator 4 end fraction


189 Views

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9.

Prove the following by using the principle of mathematical induction for all straight n element of straight N colon

space space 1 plus fraction numerator 1 over denominator 1 plus 2 end fraction plus fraction numerator 1 over denominator 1 plus 2 plus 3 end fraction plus....... plus fraction numerator 1 over denominator 1 plus 2 plus 3 plus........ plus straight n end fraction space equals space fraction numerator 2 straight n over denominator straight n plus 1 end fraction

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10.

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

1 plus 3 plus 5 plus........... space plus space left parenthesis 2 straight n minus 1 right parenthesis space equals space straight n squared


Let P(n): 1 plus 3 plus 5 plus............. plus left parenthesis 2 straight n minus 1 right parenthesis space equals space straight n squared

I.         For n = 1,

           P(1) : 1 = left parenthesis 1 right parenthesis squared rightwards double arrow space 1 space equals space 1

∴         P(1) is true

II.        Let the statement be true for n = m, straight m element of straight N

rightwards double arrow     P(m) : 1 + 3 + 5 + .................... + (2m - 1) = m2                    ...(i)

III.      For n = m + 1,

         P(m + 1) : 1 + 3 + 5 + .......... + [2 (m+1) - 1] = (m + 1)2

or      1 + 3 + 5 + ........... + (2m - 1) + (2m + 1) = (m + 1)2

        From (i),

         1 plus 3 plus 5 plus.............. space plus space left parenthesis 2 straight m space minus space 1 right parenthesis space equals space straight m squared

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rightwards double arrow    straight m squared space plus space 2 straight m space plus space 1 space equals space left parenthesis straight m space plus space 1 right parenthesis squared space rightwards double arrow space left parenthesis straight m space plus space 1 right parenthesis squared space equals space left parenthesis straight m space plus space 1 right parenthesis squared

∴       P (m + 1) is true.

∴       P(m) is true. rightwards double arrow space straight P left parenthesis straight m space plus space 1 right parenthesis is true.

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


        
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