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**Hint:**To solve the given problem one must know the formula for the sum of cubes of first ‘n’ natural numbers. Write the sum of the volumes of each cube in terms of the sum of cubes of first 30 natural numbers and sum of cubes of first 15 natural numbers.

**Formula used:**Sum of cubes of first ‘n’ natural numbers = ${{\left[ \dfrac{n(n+1)}{2} \right]}^{2}}$

**Complete Answer:**

The volume of a cube with each of length ‘a’ is equal to ${{a}^{3}}$.

Therefore, the volumes of the 15 cubes with sides 16cm, 17cm, 18cm,……. 30cm respectively are equal to ${{16}^{3}},{{17}^{3}},{{18}^{3}},{{......30}^{3}}$, respectively.

The total sum of the volumes of the 15 cubes is equal to $V={{16}^{3}}+{{17}^{3}}+{{18}^{3}}+......+{{30}^{3}}$ …. (i).

As we can see, equation (i) is a series of the cubes of 15 consecutive natural numbers starting from 16 to 30.

To the value of V, we shall use the formula for the sum of the cubes of first ‘n’ natural numbers and the formula is given as ${{\left[ \dfrac{n(n+1)}{2} \right]}^{2}}$.

Now, we can write the equation as the difference between cubes of the first 30 natural numbers and cubes of the first 15 natural numbers.

i.e. $V=\left( {{1}^{3}}+{{2}^{3}}+{{3}^{3}}+......+{{30}^{3}} \right)-\left( {{1}^{3}}+{{2}^{3}}+{{3}^{3}}+......+{{15}^{3}} \right)$ ….. (ii).

Now, for the above formula, we get that the sum of cubes of the first 30 natural numbers will be ${{\left[ \dfrac{30(30+1)}{2} \right]}^{2}}={{\left[ \dfrac{30(31)}{2} \right]}^{2}} = 216225$.

This means that $\left( {{1}^{3}}+{{2}^{3}}+{{3}^{3}}+......+{{30}^{3}} \right)=216225$.

And the sum of cubes of the first 15 natural numbers will be ${{\left[ \dfrac{15(15+1)}{2} \right]}^{2}}={{\left[ \dfrac{15(16)}{2} \right]}^{2}} = 14400$.

This means that $\left( {{1}^{3}}+{{2}^{3}}+{{3}^{3}}+......+{{15}^{3}} \right) = 14400$.

Now, substitute these values in equation (ii).

$\Rightarrow V = 216225 - 14400 = 201825c{{m}^{3}}$.

**This means that the total volume of the 15 cubes is equal to $201825c{{m}^{3}}$.**

**Note:**Note that the above formula is only valid for first consecutive natural numbers. The sum of the squares of first ‘n’ natural numbers is given as $\left[ \dfrac{n(2n+1)}{6} \right]$. The sum of first ‘n’ natural numbers is given as $\left[ \dfrac{n(n+1)}{2} \right]$.

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