A sequence $$\{a_n\}$$, $$n \geq 1$$ with $$a_1 = \frac{1}{2}$$ and $$a_n = \frac{a_{n-1}}{2na_{n-1} + 1}$$ is given. Then the value of $$a_1 + a_2 + a_3 + \ldots + a_{2024}$$ is equal to

We are given

$$a_1=\dfrac{1}{2}$$

and

$$a_n=\dfrac{a_{n-1}}{2na_{n-1}+1}.$$

We want to find

$$a_1+a_2+a_3+\cdots+a_{2024}.$$

First, take the reciprocal of both sides of the given equation:

$$\dfrac{1}{a_n}=\dfrac{2na_{n-1}+1}{a_{n-1}}.$$

Splitting the fraction,

$$\dfrac{1}{a_n}=2n+\dfrac{1}{a_{n-1}}.$$

Therefore,

$$\dfrac{1}{a_n}=\dfrac{1}{a_{n-1}}+2n.$$

Now we know that

$$a_1=\dfrac{1}{2},$$

$$\dfrac{1}{a_1}=2\times 1$$

$$\dfrac{1}{a_2}=\dfrac{1}{a_1}+2 \times 2$$

$$\dfrac{1}{a_3}=\dfrac{1}{a_2}+2 \times 3$$

$$\dfrac{1}{a_4}=\dfrac{1}{a_3}+2 \times 4$$

.

.

.

$$\dfrac{1}{a_n}=\dfrac{1}{a_{n-1}}+2 \times n$$

Adding all the equations we get

$$\dfrac{1}{a_n}=2\sum n = n(n+1)$$

Hence,

$$a_n=\dfrac{1}{n(n+1)}.$$

Now split this fraction:

$$\dfrac{1}{n(n+1)}=\dfrac{1}{n}-\dfrac{1}{n+1}.$$

Therefore,

$$a_n=\dfrac{1}{n}-\dfrac{1}{n+1}.$$

Now substitute this into the required sum:

$$a_1+a_2+a_3+\cdots+a_{2024}$$

$$=\left(1-\dfrac{1}{2}\right)+\left(\dfrac{1}{2}-\dfrac{1}{3}\right)+\left(\dfrac{1}{3}-\dfrac{1}{4}\right)+\cdots+\left(\dfrac{1}{2024}-\dfrac{1}{2025}\right).$$

Notice that most terms cancel:

$$\cancel{1-\dfrac{1}{2}}+\cancel{\dfrac{1}{2}-\dfrac{1}{3}}+\cancel{\dfrac{1}{3}-\dfrac{1}{4}}+\cdots+\left(\dfrac{1}{2024}-\dfrac{1}{2025}\right).$$

Everything in the middle cancels, leaving only

$$1-\dfrac{1}{2025}.$$

Therefore,

$$1-\dfrac1{2025}=\dfrac{2025-1}{2025}$$

$$=\dfrac{2024}{2025}$$