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find the sum of the 37th bracket of the following series

\left ( 1 \right )+\left ( 7+7^{2} \right +7^{3})+\left ( 7^{4} \right +7^{5}+7^{6}+7^{7}+7^{8})+\left ( 7^{9} \right +7^{10}+-------+7^{15})+------

  • Option 1)

    \frac{7^3^7}{6}\left ( 7^{73} -1\right )

  • Option 2)

    \frac{7^{73}-1}{6}

  • Option 3)

    7^{1260}\left ( \frac{7^{71}-1}{6}\right )

  • Option 4)

    None of these

 

Answers (1)

best_answer

As learnt in concept

Sequence -

Arragement of real numbers specified in a definite order, by some assigned law.

- wherein

Notations -

a_{1},a_{2},a_{3},a_{4}.........a_{n}

or

< a_{n}>

 

 In the 37th bracket, 

The total number of terms = 1+3+5+7...+37\: \: \: terms

= \frac{37}{2}*(2+(3n-1)2)

= 37^{2}

So the final term of bracket is 1369.

i.e. 7^{1368}

None of the options is correct. 

 


Option 1)

\frac{7^3^7}{6}\left ( 7^{73} -1\right )

Incorrect option 

Option 2)

\frac{7^{73}-1}{6}

Incorrect option 

Option 3)

7^{1260}\left ( \frac{7^{71}-1}{6}\right )

Incorrect option 

Option 4)

None of these

Correct option

Posted by

Aadil

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