Sum of the series(1.2) +(1.2+2.3) +(1.2+2.3+3.4) + ... Upto 10 brackets is equal to N, then N-1425 equals

To find the value of N in the series \( (1.2) + (1.2 + 2.3) + (1.2 + 2.3 + 3.4) + \ldots \) up to 10 brackets, we first need to understand the pattern of the series and how to calculate the sum of each bracket.

Breaking Down the Series

The series consists of terms that can be represented as follows:

• 1st term: \( 1 \cdot 2 \)
• 2nd term: \( 1 \cdot 2 + 2 \cdot 3 \)
• 3rd term: \( 1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 \)
• ... and so on, up to the 10th term.

General Formula for Each Bracket

Each bracket can be expressed as the sum of products of consecutive integers:

For the k-th term, the sum can be represented as:

Sum(k) = \( \sum_{i=1}^{k} i(i+1) \)

Calculating Each Term

To calculate \( i(i+1) \), we can simplify it:

i(i + 1) = \( i^2 + i \)

Thus, the sum becomes:

Sum(k) = \( \sum_{i=1}^{k} (i^2 + i) \)

Using the formulas for the sum of the first k integers and the sum of the squares of the first k integers:

• Sum of first k integers: \( \frac{k(k + 1)}{2} \)
• Sum of squares of first k integers: \( \frac{k(k + 1)(2k + 1)}{6} \)

Putting It All Together

Now, we can express the total sum for each k:

Sum(k) = \( \frac{k(k + 1)(2k + 1)}{6} + \frac{k(k + 1)}{2} \)

Combining these gives:

Sum(k) = \( \frac{k(k + 1)}{6} (2k + 1 + 3) = \frac{k(k + 1)(2k + 4)}{6} = \frac{k(k + 1)(k + 2)}{3} \)

Calculating N for k = 10

Now, substituting k = 10 into the formula:

N = \( \frac{10(10 + 1)(10 + 2)}{3} = \frac{10 \cdot 11 \cdot 12}{3} \)

Calculating that gives:

N = \( \frac{1320}{3} = 440 \)

Final Calculation

Now, we need to find \( N - 1425 \):

N - 1425 = 440 - 1425 = -985

Therefore, the final answer is:

-985