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If    a_{n}=\sqrt{7+\sqrt{7+\sqrt{7+......}}}    having n  radical signs then by methods of mathematical induction which is true

  • Option 1)

    a_{n}> 7,\forall \; n\geq 1\;

  • Option 2)

    \; \; a_{n}> 3,\forall \; n\geq 1\;

  • Option 3)

    \; \; a_{n}< 4,\forall \; n\geq 1\;

  • Option 4)

    \; \; a_{n}< 3,\forall \; n\geq 1

 

Answers (1)

best_answer

As we learnt in 

Steps of Mathematical Induction (Verification step) -

Step 1: Verification step

Actual verification of the proposition of the starting value n=1

- wherein

2^{3n}-1 is divisible by 7

Put n=1, It Satisfies.

 

 and

Steps of Mathematical Induction (Induction Step) -

Step 2: Induction Step

Assuming the proposition to be true for n=k, and proving it is true for value     n=k+1

-

 

 And

Steps of Mathematical Induction (Generalization Step) -

Combine Verification step and Induction step

p(1) is true and p(n) is true for n+1 assuming it is true for n

-

 

 a_{n} = \sqrt{7 +\sqrt{7 +\sqrt{7 +.....}}}

for\ n =1

a_{1} = \sqrt{7} which is less than 3 and lesser than 4

For n = k  

ak < 7

a_{k+1} = \sqrt{7 + a_{k}}

a_{k+1}^{2} = 7 + a_{k} < 7 +7 < 14

a_{k +1} < \sqrt{}14

Sq = an < 7


Option 1)

a_{n}> 7,\forall \; n\geq 1\;

This option is incorrect. 

Option 2)

\; \; a_{n}> 3,\forall \; n\geq 1\;

This option is correct. 

Option 3)

\; \; a_{n}< 4,\forall \; n\geq 1\;

This option is incorrect. 

Option 4)

\; \; a_{n}< 3,\forall \; n\geq 1

This option is incorrect. 

Posted by

divya.saini

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