Prove that 4+7+10+...+3n+1=n3n+52 for all positive integers n.

By mathematical induction, 4+7+10+...+3n+1=n3n+52 is true for all positive integers n.

Q34. - Algebra 2 | StudyQuestionHub

Q34.

Question

Prove that $4 + 7 + 10 + . . . + (3 n + 1) = \frac{n (3 n + 5)}{2}$ for all positive integers n.

Step-by-Step Solution

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Answer

By mathematical induction, $4 + 7 + 10 + . . . + (3 n + 1) = \frac{n (3 n + 5)}{2}$ is true for all positive integers n.

1Step 1. Given Information.

Given, to prove that the statement $4 + 7 + 10 + . . . + (3 n + 1) = \frac{n (3 n + 5)}{2}$ is true for all positive integers n.

2Step 2. Calculation .

When n=1, the statement is written as:

$\begin{array}{l} 4 + 7 + 10 + ... + (3 n + 1) = \frac{n (3 n + 5)}{2} \\ 4 = \frac{(1) (3 (1) + 5)}{2} \\ 4 = \frac{3 + 5}{2} \\ 4 = \frac{8}{2} \\ 4 = 4 \end{array}$

Hence the statement is true for n = 1.

Let the statement be true for n = k. Hence,

$4 + 7 + 10 + . . . + (3 k + 1) = \frac{k (3 k + 5)}{2}$

Checking the statement when n = k + 1:

$\begin{array}{l} 4 + 7 + 10 + ... + (3 n + 1) = \frac{n (3 n + 5)}{2} \\ 4 + 7 + 10 + ... + (3 (k + 1) + 1) = \frac{(k + 1) (3 (k + 1) + 5)}{2} \\ 4 + 7 + 10 + ... + (3 k + 1) + (3 k + 4) = \frac{(k + 1) (3 k + 8)}{2} \end{array}$

Plugging the value from the previous assumptions:

$\begin{array}{l} \frac{k (3 k + 5)}{2} + (3 k + 4) = \frac{(k + 1) (3 k + 8)}{2} \\ \frac{k (3 k + 5) + 2 (3 k + 4)}{2} = \frac{(k + 1) (3 k + 8)}{2} \\ \frac{3 k^{2} + 5 k + 6 k + 8}{2} = \frac{(k + 1) (3 k + 8)}{2} \\ \frac{3 k^{2} + 3 k + 8 k + 8}{2} = \frac{(k + 1) (3 k + 8)}{2} \\ \frac{3 k (k + 1) + 8 (k + 1)}{2} = \frac{(k + 1) (3 k + 8)}{2} \\ \frac{(k + 1) (3 k + 8)}{2} = \frac{(k + 1) (3 k + 8)}{2} \end{array}$