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

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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>




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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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#6 {main}</pre>.

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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


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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


Let straight P left parenthesis straight n right parenthesis colon space 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

I.      For n = 1,
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∴      P(1) is true

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

       P(m) : 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 m end fraction space equals space fraction numerator 2 straight m over denominator straight m plus 1 end fraction     ...(i)

III.   For n = m + 1,

       straight P left parenthesis straight m plus 1 right parenthesis colon space 1 space plus space fraction numerator 1 over denominator 1 plus 2 end fraction space plus space fraction numerator 1 over denominator 1 plus 2 plus 3 end fraction plus space.......... plus space fraction numerator 1 over denominator 1 plus 2 plus 3 plus........ plus left parenthesis straight m plus 1 right parenthesis end fraction space equals space fraction numerator 2 left parenthesis straight m plus 1 right parenthesis over denominator straight m plus 1 plus 1 end fraction   
 
or 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 space fraction numerator 1 over denominator 1 plus 2 plus 3 plus....... plus straight m end fraction plus fraction numerator 1 over denominator 1 plus 2 plus 3 plus....... plus left parenthesis straight m plus 1 right parenthesis end fraction space equals space fraction numerator 2 straight m plus 2 over denominator straight m plus 2 end fraction

    From (i),

      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 m end fraction space equals space fraction numerator 2 straight m over denominator straight m plus 1 end fraction 

      Also,  1, 2 , 3,..........., m+1 are in A.P.

rightwards double arrow  1 plus 2 plus 3 plus........... plus left parenthesis straight m plus 1 right parenthesis equals fraction numerator straight m plus 1 over denominator 2 end fraction open square brackets 1 plus space left parenthesis straight m plus 1 right parenthesis close square brackets space equals space fraction numerator left parenthesis straight m plus 1 right parenthesis left parenthesis straight m plus 2 right parenthesis over denominator 2 end fraction
                                                                          
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∴     space space straight P left parenthesis straight m plus 1 right parenthesis space colon space fraction numerator 2 straight m over denominator straight m plus 1 end fraction plus space fraction numerator 1 over denominator begin display style fraction numerator straight m plus 1 over denominator 2 end fraction left parenthesis straight m plus 2 right parenthesis end style end fraction space equals space fraction numerator 2 straight m space plus 2 space over denominator straight m space plus space 2 end fraction space rightwards double arrow space fraction numerator 2 straight m over denominator straight m plus 1 end fraction plus fraction numerator 2 over denominator left parenthesis straight m plus 1 right parenthesis left parenthesis straight m plus 2 right parenthesis end fraction space equals space fraction numerator 2 straight m plus 2 over denominator straight m plus 2 end fraction

rightwards double arrowfraction numerator 2 over denominator straight m plus 1 end fraction open square brackets straight m plus fraction numerator 1 over denominator straight m plus 2 end fraction close square brackets space equals space fraction numerator 2 straight m plus 2 over denominator straight m plus 2 end fraction space rightwards double arrow space fraction numerator 2 over denominator straight m plus 1 end fraction open square brackets fraction numerator straight m squared plus 2 straight m plus 1 over denominator straight m plus 2 end fraction close square brackets space equals space fraction numerator 2 straight m space plus space 2 over denominator straight m plus 2 end fraction

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         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, the statement P(n) is true for all straight n element of space straight N.



       



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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

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