Question

The sum of the geometric series \(\sqrt{3}+\sqrt{12}+\sqrt{48}+\dots\) up to \(10\) terms is

• A. \(1023\sqrt{3}\)
• B. \(1024\sqrt{3}\)
• C. \(511\sqrt{3}\)
• D. \(512\sqrt{3}\)
• E. \(215\sqrt{3}\)

Hint:

Whenever radicals appear in a sequence, first simplify them. That often reveals a standard A.P. or G.P. pattern and makes the sum much easier to calculate.

Answer 1

The Correct Option is A

Step 1: Write the given series clearly.
The series is:
\[ \sqrt{3}+\sqrt{12}+\sqrt{48}+\dots \] We must find the sum of the first \(10\) terms.

Step 2: Simplify the terms to identify the pattern.
Now simplify each radical:
\[ \sqrt{12}=\sqrt{4\cdot 3}=2\sqrt{3} \] \[ \sqrt{48}=\sqrt{16\cdot 3}=4\sqrt{3} \] So the series becomes:
\[ \sqrt{3}+2\sqrt{3}+4\sqrt{3}+\dots \]

Step 3: Identify the first term and common ratio.
From the simplified series, we see:
\[ a=\sqrt{3} \] and each term is obtained by multiplying the previous term by \(2\). Hence, the common ratio is:
\[ r=2 \] Therefore, this is a geometric progression.

Step 4: Recall the sum formula for a geometric progression.
The sum of the first \(n\) terms of a G.P. is:
\[ S_n=\frac{a(r^n-1)}{r-1}, \quad r\ne 1 \] Here,
\[ a=\sqrt{3}, \quad r=2, \quad n=10 \]

Step 5: Substitute the values into the formula.
So,
\[ S_{10}=\frac{\sqrt{3}(2^{10}-1)}{2-1} \] \[ =\sqrt{3}(2^{10}-1) \]

Step 6: Evaluate the power and simplify.
Since \[ 2^{10}=1024, \] we get:
\[ S_{10}=\sqrt{3}(1024-1)=\sqrt{3}(1023) \] Thus,
\[ S_{10}=1023\sqrt{3} \]

Step 7: Match with the given options.
The value \(1023\sqrt{3}\) matches option \((1)\). Hence, the correct answer is:
\[ \boxed{1023\sqrt{3}} \]