Question

If $2^{10}+2^9\times 3^1+2^8\times 3^2+....2\times {10}^9+3^{10}=S-2^{11}$, then $S$ is equal to

• A. $3^{11}$
• B. $\frac{3^{11}}{2}+2^{10}$
• C. $2\times 3^{11}$
• D. $3^{11}-2^{12}$

Answer 1

The correct option is A

$3^{11}$

Explanation of the correct answer:

Step 1: Assigning the sequence to a variable.

From the question,

$2^{10}+2^9\times 3^1+2^8\times 3^2+....+2\times {10}^9+3^{10}=S-2^{11}$ (i)

Let us assume the sequence be written as

$A=S-2^{11}$

where,

$A=2^{10}+2^9\times 3^1+2^8\times 3^2+....+2\times {10}^9+3^{10}$ (ii)

Step 2: Getting another equation.

On multiplying equation (ii) with $\frac{3}{2}$, we get

$\frac{3}{2}A=2^9\times 3^1+2^8\times 3^2+....+3^{10}+\frac{3^{11}}{2}$ (iii)

Step 3: Subtracting the two equations.

By subtracting the equation (ii) from (iii), we obtain

$\frac{1}{2}A=\frac{3^{11}}{2}-2^{10}$

$A=3^{11}-2^{11}$

Step 4: Substituting the value of $A$.

On substituting the value of $A$ in equation (i) we get

$3^{11}-2^{11}=S-2^{11}$

Therefore, on comparing both sides of the above equation it can be concluded that

$S=3^{11}$.

Hence, Option (A) is the correct .