Question

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

Answer 1

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

Answer 2

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`

Answer 3

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.!