Question

if 2cube+4cube+6cube+......+2n whole cube=kn square (n+1)whole square then k=

Answer 1

Answer

Answer 2

Concept

The sum of cubes of first n natural numbers is given by the formula $( \frac{n(n+1)}{2} )^{2}$

or $1^{3} + 2^{3} + 3^{3} +... + n^{3} =$ $( \frac{n(n+1)}{2} )^{2}$.

Given

$2^{3} + 4^{3} + 6^{3} +...+ 2n^{3} = kn^{2} (n+1)^{2}$

Find

we are asked to find the value of k

Solution

According to the question, we have,

$2^{3} + 4^{3} + 6^{3} +... + (2n)^{3} = kn^{2} (n+1)^{2}$

⇒ $2^{3} + (2.2)^{3} + (3.2)^{3} +... + (2n)^{3} = kn^{2} (n+1)^{2}$

⇒ $2^{3} ( 1^{3} + 2^{3} + 3^{3} +... + n^{3}) = kn^{2} (n+1)^{2}$

As we can see, after taking out $2^{3}$, the term left inside the bracket on the left-hand side is simply the sum of cubes on n natural numbers starting from 1. Thus, we can simply use the formula of the sum of cubes of n natural numbers.

⇒ $2^{3} ( \frac{n(n+1)}{2} )^{2} = kn^{2} (n+1)^{2}$

canceling $[n(n+1)]^{2}$ from both sides we get,

$2^{3} ( \frac{1}{2} )^{2} = k$

⇒ $k = \frac{8}{4}$

or, $k = 2$

Thus, the value of k in the given equation is 2.

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