25+27+29+...+(2n-1)=1456

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вопрос

To solve this problem, we need to find the value of "n" that makes the sum of the series equal to 1456.

The given series is an arithmetic series with a common difference of 2. The sum of an arithmetic series is given by the formula:

S = n/2 * (a + l)

Where:
S = Sum of the series
n = Number of terms
a = First term
l = Last term

In this case:
a = 25
l = 2n - 1
S = 1456

Plugging in the values, we get:

1456 = n/2 * (25 + 2n - 1)

1456 = n/2 * (24 + 2n)

1456 = n(24 + 2n) / 2

2912 = 24n + 2n^2

Rearranging the equation to form a quadratic equation:

2n^2 + 24n - 2912 = 0

Dividing by 2:

n^2 + 12n - 1456 = 0

Factoring the quadratic equation:

(n + 44)(n - 32) = 0

n = -44 or n = 32

Since the number of terms cannot be negative, n = 32 is the valid solution.

Therefore, the sum of the series 25 + 27 + 29 + ... + (2n - 1) equals 1456 when n = 32.