If $\mathrm{a}_{1}=\frac{1}{2 \times 5}, \mathrm{a}_{2}=\frac{1}{5 \times 8}, \mathrm{a}_{3}=\frac{1}{8 \times 11}$, then $\mathrm{a}_{1,}+\mathrm{a}_{2,}+\mathrm{a}_{3,}+\ldots . . \mathrm{a}_{100}$ is

Let's look at what we're given:

• $a_1 = \frac{1}{2 \times 5}$
• $a_2 = \frac{1}{5 \times 8}$
• $a_3 = \frac{1}{8 \times 11}$

Notice the pattern in the denominators:

• First term: 2, 5 (difference = 3)
• Second term: 5, 8 (difference = 3)
• Third term: 8, 11 (difference = 3)

The general term is:

$a_n = \frac{1}{(3n-1) \times (3n+2)}$

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Instead of trying to add these fractions directly, we use partial fractions:

$\frac{1}{(3n-1)(3n+2)} = \frac{A}{3n-1} + \frac{B}{3n+2}$

Finding A and B:

$1 = A(3n+2) + B(3n-1)$

• When $3n+2 = 0$: $1 = B(-3)$, so $B = -\frac{1}{3}$
• When $3n-1 = 0$: $1 = A(3)$, so $A = \frac{1}{3}$

Therefore:

$a_n = \frac{1}{3}\left(\frac{1}{3n-1} - \frac{1}{3n+2}\right)$

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Writing out the first few terms:

$a_1 = \frac{1}{3}\left(\frac{1}{2} - \frac{1}{5}\right)$

$a_2 = \frac{1}{3}\left(\frac{1}{5} - \frac{1}{8}\right)$

$a_3 = \frac{1}{3}\left(\frac{1}{8} - \frac{1}{11}\right)$

$a_{100} = \frac{1}{3}\left(\frac{1}{299} - \frac{1}{302}\right)$

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When we add all terms:

$\sum_{n=1}^{100} a_n = \frac{1}{3}\left[\left(\frac{1}{2} - \frac{1}{5}\right) + \left(\frac{1}{5} - \frac{1}{8}\right) + \left(\frac{1}{8} - \frac{1}{11}\right) + \cdots + \left(\frac{1}{299} - \frac{1}{302}\right)\right]$

Most terms cancel out! This is called a telescoping series.

What remains:

$\frac{1}{3}\left(\frac{1}{2} - \frac{1}{302}\right)$

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$\frac{1}{3}\left(\frac{1}{2} - \frac{1}{302}\right) = \frac{1}{3} \times \frac{302-2}{2 \times 302} = \frac{1}{3} \times \frac{300}{604}$

$= \frac{300}{3 \times 604} = \frac{100}{604} = \frac{25}{151}$

Answer: $\frac{25}{151}$