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Math  /  Algebra

Question10. For all n≥1n \geq 1, prove the following by mathematical induction: (b) 12+222+323+⋯+n2n=2−n+22n\frac{1}{2}+\frac{2}{2^{2}}+\frac{3}{2^{3}}+\cdots+\frac{n}{2^{n}}=2-\frac{n+2}{2^{n}}.

Studdy Solution

STEP 1

1. We are asked to prove the given statement for all n≥1 n \geq 1 using mathematical induction.
2. Mathematical induction involves proving a base case and an inductive step.

STEP 2

1. Prove the base case for n=1 n = 1 .
2. Assume the statement is true for n=k n = k (inductive hypothesis).
3. Prove the statement for n=k+1 n = k + 1 using the inductive hypothesis.

STEP 3

Prove the base case for n=1 n = 1 .
The left-hand side of the equation is:
12 \frac{1}{2}
The right-hand side of the equation is:
2−1+221=2−32=42−32=12 2 - \frac{1 + 2}{2^1} = 2 - \frac{3}{2} = \frac{4}{2} - \frac{3}{2} = \frac{1}{2}
Since both sides are equal, the base case holds true.

STEP 4

Assume the statement is true for n=k n = k . This is the inductive hypothesis:
12+222+⋯+k2k=2−k+22k \frac{1}{2} + \frac{2}{2^2} + \cdots + \frac{k}{2^k} = 2 - \frac{k+2}{2^k}

STEP 5

Prove the statement for n=k+1 n = k + 1 .
Consider the left-hand side for n=k+1 n = k + 1 :
12+222+⋯+k2k+k+12k+1 \frac{1}{2} + \frac{2}{2^2} + \cdots + \frac{k}{2^k} + \frac{k+1}{2^{k+1}}
Using the inductive hypothesis, substitute the known sum:
(2−k+22k)+k+12k+1 \left(2 - \frac{k+2}{2^k}\right) + \frac{k+1}{2^{k+1}}
Simplify the expression:
=2−k+22k+k+12k+1 = 2 - \frac{k+2}{2^k} + \frac{k+1}{2^{k+1}}
To combine the fractions, express k+12k+1\frac{k+1}{2^{k+1}} with a common denominator:
=2−k+22k+k+12⋅2k = 2 - \frac{k+2}{2^k} + \frac{k+1}{2 \cdot 2^k}
=2−k+22k+k+12k+1 = 2 - \frac{k+2}{2^k} + \frac{k+1}{2^{k+1}}
=2−2(k+2)+(k+1)2k+1 = 2 - \frac{2(k+2) + (k+1)}{2^{k+1}}
=2−2k+4+k+12k+1 = 2 - \frac{2k + 4 + k + 1}{2^{k+1}}
=2−3k+52k+1 = 2 - \frac{3k + 5}{2^{k+1}}
This simplifies to:
=2−(k+1)+22k+1 = 2 - \frac{(k+1) + 2}{2^{k+1}}
Thus, the statement holds for n=k+1 n = k + 1 .
Since both the base case and the inductive step have been proven, by mathematical induction, the statement is true for all n≥1 n \geq 1 .

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