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

3 Answers
Oct 5, 2017

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 #n=1#

When #n=1# the given result gives:

# 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 #n=1#.

Induction Proof - General Case

Now, Let us assume that the given result [A] is true when #n=m#, for some #m in NN, m gt 1#, in which case for this particular value of #m# we have:

# 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 #n=m+1#

Induction Proof - Summary

So, we have shown that if the given result [A] is true for #n=m#, then it is also true for #n=m+1# where #m gt 1#. But we initially showed that the given result was true for #n=1# so it must also be true for #n=2, n=3, n=4, ... # and so on.

Induction Proof - Conclusion

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

Hence we have:

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

Oct 5, 2017

Answer:

See below.

Explanation:

Considering #sum_(k=0)^n x^k = (x^(n+1)-1)/(x-1)# and

#d/dx sum_(k=0)^n x^k = sum_(k=0)^n k x^(k-1)# we have

#sum_(k=1)^n k x^k = x sum_(k=0)^n k x^(k-1) = sum_(k=0)^nkx^k = sum_(k=1)^n k x^k#

so

#sum_(k=1)^n k x^k = x d/(dx)( (x^(n+1)-1)/(x-1))=(x + (n (x-1)-1) x^(1 + n))/(x-1)^2#

Making now #x = 2# we have

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

Oct 5, 2017

Answer:

Pease refer to a Proof in the Explanation.

Explanation:

Let, #S(n)=sum_(k=1)^(k=n)k2^k.#

#:. S(n)=1*2^1+2*2^2+3*2^3+...+(n-1)2^(n-1)+n*2^n.#

#:.2S(n)=1*2^2+2*2^3+3*2^4+...+(n-1)2^n+n*2^(n+1).#

#:. S(n)-2S(n)=1*2^1+(2-1)2^2+(3-2)2^3+...+(n-ul(n-1))2^n-n*2^(n+1).#

#:. -S(n)={2^1+2^2+2^3+...+2^n}-n*2^(n+1).#

We see that, the Series in #{...}# is a GP, having #1^(st)# term

#a=2=r="common ratio,"# so that, the sum of its #n# terms is,

#{a(r^n-1)}/(r-1)={2(2^n-1)}/(2-1)=2(2^n-1).#

#:. -S(n)=2(2^n-1)-n*2^(n+1).#

#:. S(n)=n*2^(n+1)-2(2^n-1)=n*2^(n+1)-2^(n+1)+2.#

#rArr sum_(k=1)^(k=n)k2^k=(n-1)2^(n=1)+2.#

Enjoy Maths.!