The value of the sum ∑i=120 i1i+1i+1+1i+2+…+120 is

A
100
B
105
C
110
D
115

Solution:

$\begin{array}{l} \sum_{i = 1}^{20} i (\frac{1}{i} + \frac{1}{i + 1} + \frac{1}{i + 2} + \dots + \frac{1}{20}) \\ = (\frac{1}{1} + \frac{1}{2} + \dots + \frac{1}{20}) + 2 (\frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{20}) + 3 (\frac{1}{3} + \dots + \frac{1}{20}) + 20 (\frac{1}{20}) \\ = (1) (1) + \frac{1}{2} (1 + 2) + \frac{1}{3} (1 + 2 + 3) + \dots + \frac{1}{20} (1 + 2 + \dots + 20) \end{array}$ $\begin{array}{l} = \sum_{k = 1}^{20} \frac{1}{k} (1 + 2 + \dots + k) = \frac{1}{2} \sum_{k = 1}^{20} (k + 1) \\ = \frac{1}{2} \cdot \frac{1}{2} (20) (2 + 21) = 115 \end{array}$

The value of the sum $\sum_{i = 1}^{20} i [\frac{1}{i} + \frac{1}{i + 1} + \frac{1}{i + 2} + \dots + \frac{1}{20}]$ is

Solution:

Related content