# Prove that #sum_(k=1)^n k 2^k = (n-1)2^(n+1) + 2 #?

##### 3 Answers

**Induction Proof - Hypothesis**

We seek to prove that:

# S(n) = sum_(k=1)^n \ k2^k = (n-1)2^(n+1) + 2 # ..... [A]

So let us test this assertion using Mathematical Induction:

**Induction Proof - Base case:**

We will show that the given result, [A], holds for

When

# LHS = sum_(k=1)^1 \ k2^k = 1 *2^1 = 2#

# RHS = (1-1)2^(1+1) + 2 = 2 #

So the given result is **true** when

**Induction Proof - General Case**

Now, Let us **assume** that the given result [A] is true when

# sum_(k=1)^m \ k2^k = (m-1)2^(m+1) + 2 # ..... [B]

Consider the LHS of [A] with the addition of the next term, in which case we have

# LHS = sum_(k=1)^(m+1) \ k2^k #

# \ \ \ \ \ \ \ \ = {sum_(k=1)^m \ k2^k } + { (m+1)2^(m+1) } #

# \ \ \ \ \ \ \ \ = (m-1)2^(m+1) + 2 + (m+1)2^(m+1) \ \ \ # using [B]

# \ \ \ \ \ \ \ \ = {(m-1) + (m+1)}2^(m+1) + 2 #

# \ \ \ \ \ \ \ \ = (2m)2^(m+1) + 2 #

# \ \ \ \ \ \ \ \ = m2^(m+2) + 2 #

# \ \ \ \ \ \ \ \ = ((m-1)+1)2^((m+1)+1) + 2 #

Which is the given result [A] with

**Induction Proof - Summary**

So, we have shown that if the given result [A] is true for

**Induction Proof - Conclusion**

Then, by the process of mathematical induction the given result [A] is true for

Hence we have:

# sum_(k=1)^n \ k2^k = (n-1)2^(n+1) + 2 # QED

See below.

#### Explanation:

Considering

so

Making now

Pease refer to a **Proof** in the **Explanation.**

#### Explanation:

Let,

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**Enjoy Maths.!**