Hello rajeev000 -
Prove by induction. Assume that this is true for some . Then:
When you're faced with an expression like this, don't multiply out all the brackets. Look for common factors. Here we have a common factor . So factorise:
Then you can multiply out within the long bracket, and simplify. You will then get (I'll leave you to check this out):
Don't be afraid of it! You know that we're looking for an expression which is the equivalent of the original one, with replaced by . So we should have a factor now of and another one of . So take out these factors, and you get (again, I'll leave you to supply the missing lines):
All that remains is to show that the final bracket is the same as - which it is - and to show that the formula is true when . Can you do this now?