Question

If 2cube+4 cube +6 cube +...........+(2n) Cube =k nsquare (n+1) square then k=?

Answer 1

Answer:

2

Step-by-step explanation:

stepwise answers given in the picture

Answer 2

The value of k = 2

Given :

$\displaystyle \sf{ {2}^{3} + {4}^{3} + {6}^{3} + ... + {(2n)}^{3} = k {n}^{2} {(n + 1)}^{2} }$

To find :

The value of k

Formula Used :

$\displaystyle \sf{ {1}^{3} + {2}^{3} + {3}^{3} + ... + {n}^{3} } = {\bigg[ \frac{n(n + 1)}{2} \bigg]}^{2}$

Solution :

Step 1 of 2 :

Write down the given equation

Here the given equation is

$\displaystyle \sf{ {2}^{3} + {4}^{3} + {6}^{3} + ... + {(2n)}^{3} = k {n}^{2} {(n + 1)}^{2} }$

Step 2 of 2 :

Find the value of k

$\displaystyle \sf{ {2}^{3} + {4}^{3} + {6}^{3} + ... + {(2n)}^{3} = k {n}^{2} {(n + 1)}^{2} }$

$\displaystyle \sf \implies {(2 \times 1)}^{3} + {(2 \times 2)}^{3} + {(2 \times 3)}^{3} + ... + {(2 \times n)}^{3} = k {n}^{2} {(n + 1)}^{2}$

$\displaystyle \sf \implies {2}^{3}\bigg[ {1}^{3} + {2}^{3} + {3}^{3} + ... + {n}^{3}\bigg] = k {n}^{2} {(n + 1)}^{2}$

$\displaystyle \sf \implies {2}^{3} \times {\bigg[ \frac{n(n + 1)}{2} \bigg]}^{2} = k {n}^{2} {(n + 1)}^{2}$

$\displaystyle \sf \implies {2}^{3} \times \frac{{n}^{2} {(n + 1)}^{2}}{ {2}^{2} } = k {n}^{2} {(n + 1)}^{2}$

$\displaystyle \sf \implies 2{n}^{2} {(n + 1)}^{2} = k {n}^{2} {(n + 1)}^{2}$

$\displaystyle \sf \implies k{n}^{2} {(n + 1)}^{2} = 2 {n}^{2} {(n + 1)}^{2}$

$\displaystyle \sf \implies k = 2$

Hence the required value of k = 2

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