Can you prove this using mathematical induction?
#1/2+2/(2^2)+3/(2^2)+...+n/(2^n)=2-
(n+2)/(2^n)#
2 Answers
What you need to do is:
1) prove that it holds for
2) prove that, if it holds for
Explanation:
1) the first statement is easy to check, we just need to use
2) now, for the second statement, let's suppose it holds for
Therefore,
So, by the Induction Principle, the equation holds for any
Hope it helps.
I assume there is an error in the given sum and the
Induction Proof - Hypothesis
We seek to prove that:
# 1/2 + 2/2^2 + 3/2^3 + ... + n/2^n = 2-(n+2)/2^n #
Which can also be written as:
# sum_(k=1)^n \ k/2^k = 2-(n+2)/2^n # ..... [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 \ k/2^k = 1/2^1=1/2#
# RHS = 1(4-1)/3 = 2-(1+2)/2^1 = 2-3/2 1/2 #
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 \ k/2^k = 2-(m+2)/2^m # ..... [B]
Consider the LHS of [A] with the addition of the next term, in which case we have
# LHS = sum_(k=1)^(m+1) \ k/2^k #
# \ \ \ \ \ \ \ \ = (sum_(k=1)^(m) \ k/2^k) + (m+1)/2^(m+1) #
# \ \ \ \ \ \ \ \ = 2-(m+2)/2^m + (m+1)/2^(m+1) \ \ \ # using [B]
# \ \ \ \ \ \ \ \ = 2 - (2(m+2))/(2 * 2^m) + (m+1)/2^(m+1) #
# \ \ \ \ \ \ \ \ = 2 - (2m+4)/(2^(m+1)) + (m+1)/2^(m+1) #
# \ \ \ \ \ \ \ \ = 2 - ((2m+4)-(m+1))/(2^(m+1))#
# \ \ \ \ \ \ \ \ = 2 - ((2m+4-m-1))/(2^(m+1))#
# \ \ \ \ \ \ \ \ = 2 - (m+3)/(2^(m+1))#
# \ \ \ \ \ \ \ \ = 2 - ((m+1)+2)/(2^(m+1))#
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:
# 1/2 + 2/2^2 + 3/2^3 + ... + n/2^n = 2-(n+2)/2^n # QED