**Problem Statement:**4) Prove by induction on $ n $ that, for all positive integers $ n $:\[\sum_{i=1}^{n} i6^{i} = \frac{6(5n6^n - 6^n + 1)}{25}\]---**Explanation:**The problem requires using mathematical induction to prove the given equation holds for all positive integers $ n $. Here, the left side of the equation represents the summation of the series from $ i = 1 $ to $ n $ of $ i \cdot 6^i $, while the right side provides a closed form for this summation. The induction process involves:1. **Base Case:** Verify the equation for $ n = 1 $.2. **Inductive Step:** Assume the equation holds for $ n = k $ (inductive hypothesis), then prove it for $ n = k+1 $. This involves showing that:\[\sum_{i=1}^{k+1} i6^i = \frac{6(5(k+1)6^{k+1} - 6^{k+1} + 1)}{25}\]**Goal:** Demonstrate the inductive step to confirm that if the statement is true for $ n = k $, it must be true for $ n = k+1 $, thereby proving the equation for all positive integers $ n $.

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Description: **Problem Statement:**4) Prove by induction on $ n $ that, for all positive integers $ n $:\[\sum_{i=1}^{n} i6^{i} = \frac{6(5n6^n - 6^n + 1)}{25}\]---**Explanation:**The problem requires using mathematical induction to prove the given equation

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4) Prove by induction on n that, for all positive integers n: 6(5n6" – 6" + 1 i6' = %3D 25 i=

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ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

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**Problem Statement:**

4) Prove by induction on $ n $ that, for all positive integers $ n $:

\[
\sum_{i=1}^{n} i6^{i} = \frac{6(5n6^n - 6^n + 1)}{25}
\]

---

**Explanation:**

The problem requires using mathematical induction to prove the given equation holds for all positive integers $ n $. Here, the left side of the equation represents the summation of the series from $ i = 1 $ to $ n $ of $ i \cdot 6^i $, while the right side provides a closed form for this summation.

The induction process involves:

1. **Base Case:** Verify the equation for $ n = 1 $.
2. **Inductive Step:** Assume the equation holds for $ n = k $ (inductive hypothesis), then prove it for $ n = k+1 $. This involves showing that:

\[
\sum_{i=1}^{k+1} i6^i = \frac{6(5(k+1)6^{k+1} - 6^{k+1} + 1)}{25}
\]

**Goal:** Demonstrate the inductive step to confirm that if the statement is true for $ n = k $, it must be true for $ n = k+1 $, thereby proving the equation for all positive integers $ n $.

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