Change $n$ to $n-1$ in the first sum etc.

For the first step, solve an indicial equation fo find what the starting index should be. (Because of $x$ in the denominator.) It turns out to be $1$. So attempt $f(x) = \sum_{n=1}^\infty c_nx^n$.

@GEdgar Sorry, but I didn't get that 😅 It would be helpful if you elucidate upon it, since I am so new in this topic...

It should be in all ODE textbooks.

@GEdgar Actually, in our college, the course "Differential Equations " just began and it's completely new to us. It's taught before real analysis. The course began with the method to solve ODEs using power series without any preceeding introduction. So, I am here quite helpless 😅. Of course, it might be in any standard book, but just as I said, it's the problem with the course. A book begins by the introduction to differential equation and all the basic stuffs and then slowly after some chapters power series might be introduced. But, in my case due to my stated unusual problem(to others)

@GEdgar a book might be of little help or to frame it better: I don't know where this method you stated is demonstrated say, in the book Differential Equations by SL Ross. It would be so much helpful if you suggest some reference. I would be much grateful!

@GEdgar I dont want to be so arrogant so as to ask you pen down a solution to this. But if you did it, I would be indebted to you. Thank you !😊

Here's an example of Elliot's suggested procedure in practice involving a different second order ODE

@user170231 I have edited my post after following his suggestion 😅. Mind taking a look at it ?

Looks good to me! You can simplify the $n^{\rm th}$ coefficient to clean up the expression a little. For instance,$$-\frac{c_0}x-c_1+2c_2+\sum_{n\ge1}\bigg((n^2+3n+2)c_{n+2}-(n^2+n+1)c_{n+1}-nc_n\bigg)x^n=0$$where I also multiply by $-1$ on both sides just to get a positive coefficient for the leading term $c_{n+2}$

You have $2c_2+\frac {c_0}{x}+c_1 = 0$. Of course $c_0 = 0$. Then you can take $c_1$ as any value at all. Then solve for $c_2$ in terms of $c_1$.

You can't just plugin standard power series before determining if the point you are expanding around is an ordinary point or regular singular point or even irregular singular point. This should be the very first step to do in solving any ode using series methods.

Also you did not say if the expansion around $x=0$ or $x=1$?

@GEdgar How is $c_0=0$ ? Am I missing something ? 😕

@user170231 Thank you so very much for responding! But what to do next? I am literally stuck here....

@Franklin See Nasser's comment/answer, you have to be careful about applying power series method. (Although if the earlier equation was correct, $c_0=0$ follows from the right side being free of a $x^{-1}$ term.)

@user170231 Ohh..I see the reason for why $c_0=0$! So now, upon comparing the left and right side, of the equation, can we conclude $2c_2+c_1=0$ and the coefficient of $x^n$ is equal to $0$ as well ? Nasser's answer is just too much for me. I have mentioned, the reason in the comment section of Elliot Yu's answer.😅😅😅

Exactly, left side = zero, so the coefficient of each successive power of $x$ on the left should also sum to zero. (Again, assuming what was done first was correct.) You mention this equation shows up in a handout - I would consult with whomever handed it out to see if this ODE is something you're expected to be able to solve.

@user170231 The world is cruel like that 😂😂😂! Anyways, I have edited OP. But now, I need some help, solving this weird recurrence relation ?🤔

@geetha290krm I have tried to do what you said. Any suggestions how to proceed? I am stuck at the recurrence relation.( I have edited my post 😅)

This is just the introduction and nothing else Then you must have copied the wrong ode. For an intro course, the ode will have an ordinary point. Nothing more. This is the most simple case. But that is not what you showed. The ode you gave falls into the regular singular case, with one of the hardest subcases. Nothing any one can do about it. This is just how it is. If you do not like the solution, you can change the ode :)