Question

1.Find the sum of the consecutive cube upto (45)cube
2. Find the sum of the consecutive cubes upto (20)cube
3.Prove that if a number becomes half then its cube becomes one eighth of the cube of the given number.




please answer the questions​

Answer 1

Answer:

1.sum of consecutive cubes upto 'n'= ({ \frac{n(n + 1)}{2} } so,sum upto 45³ ={(45)(46)/2}² ={(45)(23)}² =(1035)²

2.Hello,

Let the number be x

When the number is cubed then it becomes

(x) ³ = x³

Now

The number is halved.

So, the new number becomes

\frac{x}{2}

2

x

Now,

When it is cubed it becomes

\begin{lgathered}{( \frac{x}{2} ) }^{3} \\ \\ = \frac{ {x}^{3} }{ {2}^{3} } = \frac{ {x}^{3} }{8}\end{lgathered}

(

2

x

)

3

=

2

3

x

3

=

8

x

3

Now

On separating we find

\frac{ {x}^{3} }{8} = {x}^{3} \times \frac{1}{8}

8

x

3

=x

3

×

8

1

So,

The cube if the new number becomes 1/8 th of the original number 's cube.

Hence,

Proved.

Answer 2

Answer:

Step-by-step explanation:

1. formula:

sum of consecutive cubes upto 'n'=

({ \frac{n(n + 1)}{2} })^{2}  

so,sum upto 45³

={(45)(46)/2}²

={(45)(23)}²

=(1035)²

=10,71,225.

3. Let the number be x

When the number is cubed then it becomes

(x) ³ = x³

Now

The number is halved.

So, the new number becomes

Now,

When it is cubed it becomes

Now

On separating we find

So,

The cube if the new number becomes 1/8 th of the original number 's cube.

Hence,

Proved.