(a) Find the sum of all the terms of the geometric sequence 1, 2, 4, ••., 64.
(b) Let a be a constant, where a > 0. (i)
Simplify a x a^2x a^4 x ... x a ^65
(i) Simplify log a + log a ^2+ log a ^4+ ... + log a ^64

### (a) Find the sum of all the terms of the geometric sequence \( 1, 2, 4, \dots, 64 \):

This is a geometric sequence where the first term \( a_1 = 1 \) and the common ratio \( r = 2 \). The last term is \( 64 \), so we need to find the sum of the sequence.

The formula for the sum of the first \( n \) terms of a geometric sequence is:

\[
S_n = \frac{a_1 (1 - r^n)}{1 - r}
\]

To determine \( n \), we use the formula for the \( n \)-th term of a geometric sequence:

\[
a_n = a_1 r^{n-1}
\]

Given \( a_n = 64 \), we solve for \( n \):

\[
64 = 1 \cdot 2^{n-1}
\]
\[
64 = 2^{n-1}
\]
\[
n-1 = 6 \quad \Rightarrow \quad n = 7
\]

Now we can find the sum:

\[
S_7 = \frac{1 (1 - 2^7)}{1 - 2} = \frac{1 (1 - 128)}{-1} = \frac{-127}{-1} = 127
\]

Thus, the sum of all the terms of the geometric sequence is:

\[
\boxed{127}
\]

---

### (b) (i) Simplify \( a \times a^2 \times a^4 \times \dots \times a^{64} \):

This is a product of powers of \( a \). To simplify, we add the exponents:

\[
a \times a^2 \times a^4 \times \dots \times a^{64} = a^{1 + 2 + 4 + \dots + 64}
\]

This is a sum of powers of 2: \( 1 + 2 + 4 + 8 + 16 + 32 + 64 \).

This sum is a geometric series with \( a_1 = 1 \), \( r = 2 \), and the number of terms \( n = 7 \). The sum of the exponents is:

\[
S = \frac{1(2^7 - 1)}{2 - 1} = 2^7 - 1 = 128 - 1 = 127
\]

Thus, the simplified expression is:

\[
a^{127}
\]

---

### (b) (ii) Simplify \( \log a + \log a^2 + \log a^4 + \dots + \log a^{64} \):

Using the logarithm property \( \log(x^y) = y \log(x) \), we can rewrite each term:

\[
\log a^1 + \log a^2 + \log a^4 + \dots + \log a^{64} = 1 \log a + 2 \log a + 4 \log a + \dots + 64 \log a
\]

Factor out \( \log a \):

\[
\log a (1 + 2 + 4 + \dots + 64)
\]

We already calculated the sum \( 1 + 2 + 4 + \dots + 64 = 127 \), so:

\[
127 \log a
\]

Thus, the simplified expression is:

\[
\boxed{127 \log a}
\]

---

Would you like further details on any part of the solution? Here are some related questions for more exploration:

1. How do you find the sum of the first \( n \) terms of an arithmetic series?
2. What is the general formula for the sum of an infinite geometric series?
3. How would the sum change if the common ratio was less than 1?
4. How can you express the product of terms in a geometric progression using logarithms?
5. Can you explain how logarithmic identities simplify complex expressions?

**Tip:** Always remember that geometric sequences grow exponentially, and logarithms help simplify exponential expressions.

(a) Find the sum of all the terms of the geometric sequence 1, 2, 4, ••., 64.
(b) Let a be a constant, where a > 0. (i) Simplify a x a^2 x a^4 x ... x a ^65
(ii) Simplify log a + log a ^2 + log a ^4 + ... + log a ^64

Learn how to find the sum of a geometric sequence (1, 2, 4, ..., 64) and simplify both exponential and logarithmic expressions step-by-step. This detailed explanation covers key concepts in geometric sequences, exponents, and logarithms. Suitable for students in Grades 10-12.

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