Proof by Mathematical induction, ∑r^i = (r^(n+1)-1)/(r-1) for r≠0,r ≠1,n∈N. Summation

prove by mathematical induction, summation of r^i from i=0 to n= (r^(n+1)-1)/(r-1) for r≠0,r ≠1,n∈N. calculus.

[CARBON-AI]

Base Case (n = 0):
The statement is true for n = 0 as
∑(r^i) = r^0 = 1
and (r^(0+1) - 1)/(r-1) = (r^1 - 1)/(r-1) = 1

Inductive Step:
Assume that the statement is true for n = k, i.e.
∑(r^i) = (r^(k+1) - 1)/(r-1)

Now we need to prove that the statement is also true for n = k+1
∑(r^i) = (r^0 + r^1 + r^2 + ... + r^k) + r^(k+1)
= (r^(k+1) - 1)/(r-1) + r^(k+1) (by the assumption)
= (r^(k+1)(r-1) + r^(k+1))/(r-1)
= (r^(k+2) - 1)/(r-1)

Therefore, the statement is true for n = k+1, if it is true for n = k.

Since the base case is true and the inductive step is true for all n, by mathematical induction, the statement is true for all n.

So, for any value of r≠0, r ≠1 and n∈N, the summation of r^i from i=0 to n= (r^(n+1)-1)/(r-1)

[erme]

If the number was 1 we just do exponent+1 and ignore the base because 1 exponents always 1, for example 1^0+1^1………1^5 = last numbers exponent +1, 5+1=6

[tesfukutapa]

find the inverse of the function x+sinx