Monday, April 15, 2019

sequences and series - Tricky question on binomial

Let's say there's a series of the form S=1102+1312104+135123106+... Now i had written the rth term as Tr=135....(2r1)123....r102r=2r!r!r!2r102r I came to the second equivalence by mutliplying and dividing the first expression with 246....2rand then taking out a power of 2 from each of the even numbers multiplied in the denomininator. From the looks of it, these expressions tend to give the idea of being solved using binomial most probably the expansion for negative indices but I don't understand how to get to the result from here

No comments:

Post a Comment

analysis - Injection, making bijection

I have injection f:AB and I want to get bijection. Can I just resting codomain to f(A)? I know that every function i...