Question

The sum $1^2-2 \cdot 3^2+3 \cdot 5^2-4 \cdot 7^2+5 \cdot 9^2-\ldots+15 \cdot 29^2$ is_____

Hint:

When faced with alternating sums, separate the odd-placed and even-placed terms, then apply the relevant summation formulas to simplify the calculations.

Answer 1

Correct Answer: 6952

Separating odd placed and even placed terms we get
$S=\left(1.1^2+3.5^2+\dots .15.(29{)}^2\right)-(2.3^2+4.7^2+\dots .+14.(27{)}^2$
$S=\sum _{n=1}^8(2n-1)(4n-3{)}^2-\sum _{n=1}^7(2n)(4n-1{)}^2$
Applying summation formula we get
$=29856-22904=6952$
So, the correct answer is 6952.

Answer 2

We are given the sum: \[ S = 1^2 - 2 \cdot 3^2 + 3.5^2 - 4.7^2 + 5.9^2 - \dots + 15.29^2. \] First, separate the odd-placed and even-placed terms: \[ S = (1^2 + 3.5^2 + \dots + 15.29^2) - (2^2 + 4.7^2 + \dots + 14.27^2). \] Step 1: We can express the sum as: \[ S = \sum_{n=1}^{8} (2n-1)^2 \cdot (4n-3)^2 - \sum_{n=1}^{7} (2n)(4n-1)^2. \] Step 2: Now, apply the summation formula: \[ S = \sum_{n=1}^{8} (2n - 1)(4n-3)^2 - \sum_{n=1}^{7} (2n)(4n-1)^2 = 29856 - 22904. \] Thus, the sum is: \[ S = 6952. \]