Deriving the Closed-Form of a Summation

Deriving the Closed-Form of \( S(n) = \sum_{k=0}^n k2^k \)

Exploring the Perturbation Method

Step 1

Adding the Last Term

We start by expressing \( S(n+1) \) by adding the next term in the sequence to the existing sum \( S(n) \). This provides our first perspective on the expanded sum.

\[ S(n+1) = S(n) + (n+1) 2^{n+1} \]

Step 2

Extracting the First Term

Alternatively we pull out the very first term where the index is zero. We then examine the remaining summation from \( k=1 \) up to \( k=n+1 \).

\[ S(n+1) = 0 \cdot 2^0 + \sum_{k=1}^{n+1} k 2^k \]

Step 3

Re-indexing and Substitution

We shift the index so it matches the limits of the original sum. This allows us to expand the terms and identify a standard geometric series component.

\begin{align*} S(n+1) &= \sum_{k=0}^n (k+1)2^{k+1} \\ &= 2 \sum_{k=0}^n (k+1)2^k \\ &= 2 \left( \sum_{k=0}^n k2^k + \sum_{k=0}^n 2^k \right) \\ &= 2(S(n) + 2^{n+1} - 1) \end{align*}

Step 4

Final Algebraic Solution

We equate the two expressions for the expanded sum and perform the final isolation of the unknown variable.

\begin{align*} S(n) + (n+1)2^{n+1} &= 2S(n) + 2 \cdot 2^{n+1} - 2 \\ S(n) &= n \cdot 2^{n+1} - 2^{n+1} + 2 \\ S(n) &= (n-1)2^{n+1} + 2 \end{align*}

Extended Verification

For \( n=0 \): Manual sum is \( 0 \cdot 2^0 = 0 \). Formula gives \( (0-1)2^1 + 2 = -2 + 2 = 0 \).

For \( n=1 \): Manual sum is \( 0 + 1 \cdot 2^1 = 2 \). Formula gives \( (1-1)2^2 + 2 = 0 + 2 = 2 \).

For \( n=3 \): Manual sum is \( 10 + 3 \cdot 2^3 = 10 + 24 = 34 \). Formula gives \( (3-1)2^4 + 2 = 2 \cdot 16 + 2 = 34 \).

For \( n=4 \): Manual sum is \( 34 + 4 \cdot 2^4 = 34 + 64 = 98 \). Formula gives \( (4-1)2^5 + 2 = 3 \cdot 32 + 2 = 98 \).

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