Question

the sum of the squares of the first n natural number is 285 while the sum of their cube is 2025 find the value of n​

Answer 1

Answer:

The value of n is 9.

Step-by-step explanation:

Consider the provided information.

The formula for sum of square of the first n natural number is:

$\frac{n(n+1)(2n+1)}{6}$

The formula for sum of cube of the first n natural number is:

$[\frac{n(n+1)}{2}]^2$

As it is given that the sum of the squares of the first n natural number is 285, therefore.

$\frac{n(n+1)(2n+1)}{6}=285$

It is given that the sum of their cube is 2025, therefore.

$[\frac{n(n+1)}{2}]^2=2025$

Now, take the square root.

$\frac{n(n+1)}{2}=\pm45$

Ignore the negative number as the number should be natural.

$\frac{n(n+1)}{2}=45$

$n(n+1)=90$

Substitute the value of $n(n+1)=90$ in$\frac{n(n+1)(2n+1)}{6}=285$.

$\frac{90(2n+1)}{6}=285$

$15(2n+1)=285$

$30n+15=285$

$30n=270$

$n=9$

Thus, the value of n is 9.

Answer 2

Answer:

The value of n is 9.

Step-by-step explanation:

Consider the provided information.

The formula for sum of square of the first n natural number is:

\frac{n(n+1)(2n+1)}{6}

6

n(n+1)(2n+1)

The formula for sum of cube of the first n natural number is:

[\frac{n(n+1)}{2}]^2[

2

n(n+1)

]

2

As it is given that the sum of the squares of the first n natural number is 285, therefore.

\frac{n(n+1)(2n+1)}{6}=285

6

n(n+1)(2n+1)

=285

It is given that the sum of their cube is 2025, therefore.

[\frac{n(n+1)}{2}]^2=2025[

2

n(n+1)

]

2

=2025

Now, take the square root.

\frac{n(n+1)}{2}=\pm45

2

n(n+1)

=±45

Ignore the negative number as the number should be natural.

\frac{n(n+1)}{2}=45

2

n(n+1)

=45

n(n+1)=90n(n+1)=90

Substitute the value of n(n+1)=90n(n+1)=90 in\frac{n(n+1)(2n+1)}{6}=285

6

n(n+1)(2n+1)

=285 .

\frac{90(2n+1)}{6}=285

6

90(2n+1)

=285

15(2n+1)=28515(2n+1)=285

30n+15=28530n+15=285

30n=27030n=270

n=9n=9