To solve the problem of finding the sum of the series (1 + (1+2) + (1+2+3) + \dots + (1+2+\dots+10)), follow these steps:

Step 1: Recognize the pattern

Each term in the series is a triangular number. The (k)-th triangular number is given by:
(T_k = \frac{k(k+1)}{2})

The series sum is the sum of the first 10 triangular numbers:
(S = \sum_{k=1}^{10} Tk = \sum{k=1}^{10} \frac{k(k+1)}{2})

Step 2: Simplify the sum

(S = \frac{1}{2} \sum_{k=1}^{10} (k^2 + k))

We use the formulas for sum of first (n) natural numbers and squares:
(\sum{k=1}^n k = \frac{n(n+1)}{2})
(\sum{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6})

Step 3: Compute for (n=10)

For (n=10):
(\sum{k=1}^{10} k = \frac{10 \times 11}{2} = 55)
(\sum{k=1}^{10} k^2 = \frac{10 \times 11 \times 21}{6} = 385)

Add these: (55 + 385 = 440)

Multiply by (\frac{1}{2}): (S = \frac{1}{2} \times 440 = 220)

Answer: (\boxed{220})

作者声明:本文包含人工智能生成内容。