The sum of the following infinite series $frac{4}{20}+frac{4cdot 7}{20cdot 30}+frac{4cdot 7cdot 10}{20cdot 30 cdot 40}+cdots$

The sum of the following infinite series $\displaystyle \frac{4}{20}+\frac{4\cdot 7}{20\cdot 30}+\frac{4\cdot 7\cdot 10}{20\cdot 30 \cdot 40}+\cdots$

$\bf{My\; Try::}$ We can write the given series as $$\left(1+\frac{4}{20}+\frac{4\cdot 7}{20\cdot 30}+\frac{4\cdot 7\cdot 10}{20\cdot 30 \cdot 40}+\cdots\right)-1$$

Now camparing with $$(1+x)^n = 1+nx+\frac{n(n-1)x^2}{2!}+\cdots$$

So we get $\displaystyle nx=\frac{4}{20}$ and $\displaystyle \frac{n(n-1)x^2}{2}=\frac{4\cdot 7}{20\cdot 30}$

So we get $$\frac{nx\cdot (nx-x)}{2}=\frac{4\cdot 7}{20\cdot 30}\Rightarrow \frac{4}{20}\cdot \left(\frac{4-20}{20}\right)\cdot \frac{1}{2}x^2=\frac{4}{20}\cdot \frac{7}{30}$$

But here $x^2=\text{Negative.}$

I did not understand how can I solve it

Help me, Thanks

Answer

Since the question asks about $X=\frac4{20}(1+\frac7{30}(1+...)))$, consider
$$1+\frac1{10}(1+\frac4{20}(1+\frac7{30}(...)))$$
The $10,20,30$ have a factor $1,2,3$ which will become $n!$ in the denominator.
Then $\frac1{10},\frac4{10},\frac7{10}$ increase by $3/10$ each time. Take the factor $3/10$ out, and we have $\frac13,\frac43,\frac73$ which increase by 1 each time. Let $x=3/10,n=1/3$.
$$1+xn+\frac{x^2}{2!}n(n+1)+\frac{x^3}{3!}n(n+1)(n+2)+...\\=(1-x)^{-n}=0.7^{-1/3}$$
This equals $1+\frac1{10}(1+X)$,
so your sum is $X=10(0.7)^{-1/3}-11$