Question

4 + 6 + 9 + 13 + 18 + ...

Answer 1

Let $$T_n$$  be the nth term and  $$S_n$$ be the sum of n terms of the given series. 

Thus, we have:

$$S_n=4+6+9+13+18+...+T_{n-1}+T_n$$ ...(1)

Equation (1) can be rewritten as:

$$S_n=4+6+9+13+18+...+T_{n-1}+T_n$$  ...(2)

On subtracting (2) from (1), we get:

$$S_n=4+6+9+13+18+...+T_{n-1}+T_n$$

$$S_n=4+6+9+13+18+...+T_{n-1}+T_n$$

$$0=4+[2+3+4+5+6+...+(T_n-T_{n-1})]-T_n$$

The sequence of difference between successive terms is 2, 3, 4, 5,...
We observe that it is an AP with common difference 1 and first term 2.
Now,

$$4+\left[\frac{(n-1)}{2}\{4+(n-2)1\}\right]-T_n=0$$

$$\Rightarrow 4+\left[\frac{(n-1)}{2}(n+2)\right]-T_n=0$$

$$\Rightarrow 4+[\frac{n^2+n}{2}-1]-T_n=0$$

$$\Rightarrow [\frac{n^2}{2}+\frac{n}{2}+3]=T_n$$

$$\because S_n=\sum _{k=1}^n{T}_k$$

$$\therefore S_n=\sum _{k=1}^n(\frac{k^2}{2}+\frac{k}{2}+3)$$

$$=\frac{1}{2}\sum _{k=1}^n{k}^2+\frac{1}{2}\sum _{k=1}^{n}k+\sum _{k=1}^{n}3$$

$$=\frac{n(n+1)(2n+1)}{2\times 6}+\frac{n(n+1)}{2\times 2}+3n$$

$$=n\left(\frac{2n^2+3n+1+3n+3+36}{12}\right)$$

$$=\frac{n}{12}(2n^2+6n+40)$$

$$=\frac{n}{6}(n^2+3n+20)$$