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Can someone explain - Sequence and series - JEE Main-2

1^{3}-2^{3}+3^{3}-4^{3}+.........+9^{3}=

  • Option 1)

    425

  • Option 2)

    -425

  • Option 3)

    475

  • Option 4)

    -475

 
Answers (1)
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As we learnt in 

Summation of series of natural numbers -

\sum_{k=1}^{n}K^{3}=\left ( \sum_{k=1}^{n}K \right )^{2}

- wherein

1^{3}+2^{3}+3^{3}+-------n^{3}\\= \left ( 1+2+3+-----+n \right )^{2}

 

1^{3}-2^{3}+3^{3}-4^{3}+\cdot \cdot +9^{3}= ? 

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

= \left ( 1^{3}+3^{3}+5^{3}+7^{3}+9^{3} \right )-8\left ( 1^{3}+2^{3}+3^{5}+4^{3} \right )        - (i)

Now we know that

1^{3}+2^{3}+3^{3}+4^{3}+\cdot \cdot 9^{3}=\left ( \frac{9\times 2}{2} \right )^{2}=\left ( 45 \right )^{2}

\left [ \because 1^{3}+2^{3}+\cdot \cdot n^{3} = \left ( \frac{n\left ( n+1 \right )^{2}}{2} \right )\right ]

So that

\left ( 1^{3}+3^{3}+5^{3}+7^{3}+9^{3} \right )= \left ( 45 \right )^{2}-8\left ( 1^{3}+2^{3}+3^{3}+4^{3} \right )

Put in (i)

\therefore \left ( 45 \right )^{2}-8\left ( 1^{3}+2^{3}+3^{3}+4^{3} \right )-8\left ( 1^{3}+2^{3}+3^{3}+4^{3} \right )

=\left ( 45 \right )^{2}-16\left ( 1^{3}+2^{3}+3^{3}+4^{3} \right )

= \left ( 45 \right )^{2}-16\times 100= \left ( 45 \right )^{2}- 1600

= \left ( 45 \right )^{2}-\left ( 40 \right )^{2}

= \left ( 45-40 \right )\left ( 45+40 \right )

= 5\times 85

= 425


Option 1)

425

This option is correct.

Option 2)

-425

This option is incorrect.

Option 3)

475

This option is incorrect.

Option 4)

-475

This option is incorrect.

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