Calculus Questions

Question

Apply mathematical induction to prove the statement. Please show the work and steps taken.

4 + 4^2 + 4^3 + . . . + 4^n = 4/3 (4n - 1)

Answer 1

Clarissa, The problem isn’t presented accurately.

S(n) = (4/3)*(4ⁿ -1) works.

Does that help?

The way to prove by induction is to do two things:

(a) demonstrate that the formula works for n = 1

(b) demonstrate algebraically that if we assume it works for n, then when you add the next term, the formula works for n +1.

So do (a):, for n = 1

S(1) = (4/3) *(4¹–1)= (4/3) * (4-1) = (4/3) * 3 = 4

Now do (b): assuming it works for n, add the (n+1)th term.

(Start): S(n+1) = (4/3)*(4ⁿ -1) + 4ⁿ⁺¹

now manipulate that until you prove that

(Finish): S(n+1) = (4/3)*(4ⁿ⁺¹ -1)

I would start by multiplying out the parentheses of (Start)

(4/3)*(4ⁿ -1) + 4ⁿ⁺¹

= (4/3)*4ⁿ - (4/3) + 4ⁿ⁺¹

= 4ⁿ⁺¹/3 - 4/3 + 4ⁿ⁺¹

Now combine like terms, specifically the 4ⁿ⁺¹ terms

= 4ⁿ⁺¹ (1+1/3) -4/3

= 4ⁿ⁺¹ (4/3) -4/3

now factor out 4/3

= (4/3)(4ⁿ⁺¹ -1)

Notice that does equal (Finish)