Question

• Part 1 : Prove by induction that $\sum_{i=0}^{k} i2^i = (k - 1)2^{k+1} + 2$ • Part 2 : Prove that $\sum_{i=1}^{n} \log(i) = \Theta(n \log(n))$ • Part 3 : What is $\sum_{i=0}^{\log_2(n)} 8^i$ equal to in ?-notation? (No formal proof necessary, just a brief explanation.) HINT: use the formula for geometric sum: $\sum_{i=0}^{k} r^i = (r^{k+1} - 1)/(r - 1)$. This is generally a very useful formula.

          • Part 1 : Prove by induction that $\sum_{i=0}^{k} i2^i = (k - 1)2^{k+1} + 2$
• Part 2 : Prove that $\sum_{i=1}^{n} \log(i) = \Theta(n \log(n))$
• Part 3 : What is $\sum_{i=0}^{\log_2(n)} 8^i$ equal to in ?-notation? (No
formal proof necessary, just a brief explanation.)
HINT: use the formula for geometric sum: $\sum_{i=0}^{k} r^i = (r^{k+1} - 1)/(r - 1)$. 
This is generally a very useful formula.

• Part 1 : Prove by induction that ∑i=0^k i2^i = (k - 1)2^k+1 + 2
• Part 2 : Prove that ∑i=1^nlog(i) = Θ(n log(n))
• Part 3 : What is ∑i=0^log2(n) 8^i equal to in ?-notation? (No
formal proof necessary, just a brief explanation.)
HINT: use the formula for geometric sum: ∑i=0^k r^i = (r^k+1 - 1)/(r - 1).
This is generally a very useful formula.

Added by Amanda W.

Computer Science and Information Technology

Trishna Knowledge Systems 2018 Edition

Instant Answer

Solved by Expert Jennifer Hudspeth

Step 1

Base case (k = 1): When k = 1, the formula becomes E = Σi^2i = (1 - 1)^2(1+1) + 2 = 0^2(2) + 2 = 0 + 2 = 2. So the formula holds for the base case. Inductive step: Assume that the formula holds for some arbitrary value of k, i.e., E = Σi^2i = (k - 1)^2k+1 + Show more…

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Please answer me ASAP. Problem 1. Part 1: Prove by induction that E = Î£i^2i = (k - 1)^2k+1 + 2. Part 2: Prove that Ln = 1 log(i) = O(n log(n)). Part 3: (No formal proof necessary, just a brief explanation.) This is generally a very useful formula.