Prove that the sum of two consecutive powers of the number 5 is divisible by 30. If two consecutive powers are \(5^n\) and \(5^{n+1}\), then their sum can be expressed as \(k \cdot 30\), where \(k\) is an integer.

To prove that the sum of two consecutive powers of 5 is divisible by 30, let's consider two consecutive powers of 5, which are \(5^n\) and \(5^{n+1}\).

The sum of these two powers is: \(5^n + 5^{n+1}\)

We can factor out the common factor \(5^n\) from both terms: \(5^n (1 + 5)\)

Now, simplifying the term inside the parentheses gives: \(5^n \cdot 6\)

Since \(5^n\) is an exponential expression with a base of 5 and an integer exponent \(n\), it is always going to be a multiple of 5. The other factor, 6, is a multiple of 3 (and 2).

Because \(5^n \cdot 6\) is a product of a multiple of 5 and a multiple of 3, any number that is a multiple of both 5 and 3 will also be a multiple of their least common multiple (LCM). The LCM of 5 and 3 is 15.

Moreover, since we have a factor of 2 as well in the number 6, it ensures the product is also a multiple of \(2 \times 15 = 30\).

Thus, the sum \(5^n \cdot 6\) is indeed a multiple of 30, which proves that the expression \(k \cdot 30\), where \(k\) is an integer, correctly represents the sum of two consecutive powers of 5.