Let P(n) be the given statement. Now, P(n) #160;=1+3+32+...+3n 1=3n 12Step #160;1:P(1) #160;=1 #160;=31 12=22=1Hence, #160;P(1) #160;is #160;true. ...

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Statement

1+3+32+…+3n-1...

Question

1 + 3 + 3² + ... + 3ⁿ⁻¹ = $\frac{3^{n} - 1}{2}$

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Solution

Let P(n) be the given statement.
Now,
$P (n) = 1 + 3 + 3^{2} + . . . + 3^{n - 1} = \frac{3^{n} - 1}{2} Step 1 : P (1) = 1 = \frac{3^{1} - 1}{2} = \frac{2}{2} = 1 Hence, P (1) is true . Step 2 : Let P (m) is true . Then, 1 + 3 + 3^{2} + . . . + 3^{m - 1} = \frac{3^{m} - 1}{2} W e s h a l l p r o v e t h a t P (m + 1) is true . T h a t i s, 1 + 3 + 3^{2} + . . . + 3^{m} = \frac{3^{m + 1} - 1}{2} Now, w e h a v e : 1 + 3 + 3^{2} + . . . + 3^{m - 1} = \frac{3^{m} - 1}{2} \Rightarrow 1 + 3 + 3^{2} + . . . + 3^{m - 1} + 3^{m} = \frac{3^{m} - 1}{2} + 3^{m} [Adding 3^{m} to both sides] \Rightarrow 1 + 3 + 3^{2} + . . . + 3^{m} = \frac{3^{m} - 1 + 2 \times 3^{m}}{2} = \frac{3^{m} (1 + 2) - 1}{2} = \frac{3^{m + 1} - 1}{2} Hence, P (m + 1) is true . B y t h e p rinciple of m athematical i nduction, P (n) is true for all n \in N .$

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\frac{1 \times 2^{2} + 2 \times 3^{2} + . . . + n \times {(n + 1)}^{2}}{1^{2} \times 2 + 2^{2} \times 3 + . . . + n^{2} \times (n + 1)} = \frac{3 n + 5}{3 n + 1}

Show that

$\frac{1 \times 2^{2} + 2 \times 3^{2} + \dots \dots + n \times (n + 1)^{2}}{1^{2} \times 2 + 2^{2} \times 3 + \dots \dots + n^{2} \times (n + 1)} = \frac{3 n + 5}{3 n + 1}$

$1 + 3 + 3^{2} + \dots + 3^{n - 1} = \frac{(3^{n} - 1)}{2}$

Q. Show that

1 + 3 + 3^{2} + . . . . . 3^{n - 1} = \frac{3^{n} - 1}{2}

Q. Prove that :

1 + 3 + 3^{2} + . . . + 3^{n - 1} = \frac{(3^{n} - 1)}{2}

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1 + 3 + 32 + ... + 3n−1 = 3n-12

1 + 3 + 3² + ... + 3ⁿ⁻¹ = $\frac{3^{n} - 1}{2}$