Question

5+9+13+...+(4n +1)=324

Answer 1

Ответ равен 74 вставляется в место точек

Answer 2

To solve the equation 5 + 9 + 13 + ... + (4n + 1) = 324, we need to find the value of 'n' that satisfies this equation.

The given series is an arithmetic series with a common difference of 4, starting with the first term 5 and ending with the term (4n + 1). We can use the formula for the sum of an arithmetic series to find the value of 'n'.

The formula for the sum of an arithmetic series is: Sum = (n/2) * (first term + last term)

In this case, the first term (a) is 5, and the last term (l) is (4n + 1). So, the sum is 324:

324 = (n/2) * (5 + 4n + 1)

Now, let's simplify the equation:

324 = (n/2) * (4n + 6)

Next, let's get rid of the fraction by multiplying both sides by 2:

648 = n * (4n + 6)

Now, we have a quadratic equation. Let's rearrange it to be in standard form:

4n^2 + 6n - 648 = 0

To solve this quadratic equation, we can use factoring, completing the square, or the quadratic formula. In this case, factoring is the easiest method:

4n^2 + 6n - 648 = (4n + 54)(n - 12) = 0

Now, set each factor to zero and solve for 'n':

1. 4n + 54 = 0 4n = -54 n = -54/4 n = -13.5

2. n - 12 = 0 n = 12

We obtain two solutions for 'n': n = -13.5 and n = 12.

However, in this context, it doesn't make sense to have a negative number of terms in the series. Thus, the valid solution for 'n' is n = 12.

So, the value of 'n' that satisfies the equation is 12.