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

Option 1)
$425$

Option 2)
$-425$

Option 3)
$475$

Option 4)
$-475$

Answers (2)

best_answer

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.

Posted by

Plabita

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425

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