Let $$A_n = \left(\frac{3}{4}\right) - \left(\frac{3}{4}\right)^2 + \left(\frac{3}{4}\right)^3 - \ldots + (-1)^{n-1}\left(\frac{3}{4}\right)^n$$ and $$B_n = 1 - A_n$$. Then, the least odd natural number p, so that $$B_n > A_n$$, for all $$n \geq p$$ is:

We have the finite alternating geometric sum

$$A_n=\left(\frac34\right)-\left(\frac34\right)^2+\left(\frac34\right)^3-\ldots+(-1)^{\,n-1}\left(\frac34\right)^n.$$

The first term is $$a=\dfrac34$$ and the common ratio is $$r=-\dfrac34.$$

For any geometric progression the sum of the first $$n$$ terms is given by the well-known formula

$$S_n=\dfrac{a\,(1-r^{\,n})}{1-r}.$$

Substituting $$a=\dfrac34$$ and $$r=-\dfrac34$$, we get

$$A_n=\dfrac{\dfrac34\bigl(1-(-\dfrac34)^{\,n}\bigr)}{1-(-\dfrac34)} =\dfrac{\dfrac34\bigl(1-(-\dfrac34)^{\,n}\bigr)}{1+\dfrac34} =\dfrac{\dfrac34\bigl(1-(-\dfrac34)^{\,n}\bigr)}{\dfrac74} =\dfrac34\cdot\dfrac47\bigl(1-(-\tfrac34)^{\,n}\bigr).$$

Simplifying the product $$\dfrac34\cdot\dfrac47$$ we obtain

$$A_n=\dfrac37\bigl(1-(-\tfrac34)^{\,n}\bigr).$$

Now, by definition,

$$B_n=1-A_n =1-\dfrac37\bigl(1-(-\tfrac34)^{\,n}\bigr) =1-\dfrac37+\dfrac37(-\tfrac34)^{\,n} =\dfrac47+\dfrac37\bigl(-\tfrac34\bigr)^{\,n}.$$

We want $$B_n>A_n.$$ Substituting the above expressions gives

$$\dfrac47+\dfrac37\bigl(-\tfrac34\bigr)^{\,n}\;>\;\dfrac37\bigl(1-(-\tfrac34)^{\,n}\bigr).$$

Multiplying every term by 7 to clear denominators,

$$4+3\bigl(-\tfrac34\bigr)^{\,n}\;>\;3\Bigl[1-(-\tfrac34)^{\,n}\Bigr].$$

Expanding the right-hand side,

$$4+3\bigl(-\tfrac34\bigr)^{\,n}\;>\;3-3\bigl(-\tfrac34\bigr)^{\,n}.$$

Bringing every term to the left,

$$4+3\bigl(-\tfrac34\bigr)^{\,n}-3+3\bigl(-\tfrac34\bigr)^{\,n}\;>\;0,$$

which simplifies to

$$1+6\bigl(-\tfrac34\bigr)^{\,n}\;>\;0.$$

So the inequality $$B_n>A_n$$ is equivalent to

$$1+6(-\tfrac34)^{\,n}>0.$$

Notice that $$(-\tfrac34)^{\,n}$$ alternates its sign:

• If $$n$$ is even, $$(-\tfrac34)^{\,n}=(\tfrac34)^{\,n}>0$$, hence $$1+6(\text{positive})>0$$ is always true.
• If $$n$$ is odd, $$(-\tfrac34)^{\,n}=-(\tfrac34)^{\,n}$$, and the condition becomes
  $$1-6\bigl(\tfrac34\bigr)^{\,n}>0 \quad\Longrightarrow\quad \bigl(\tfrac34\bigr)^{\,n}<\dfrac16.$$

Because $$\bigl(\tfrac34\bigr)^{\,n}$$ decreases as $$n$$ increases, once it falls below $$\dfrac16$$ it will stay below that value for all larger $$n$$. We therefore search for the smallest odd natural number $$n$$ satisfying

$$\bigl(\tfrac34\bigr)^{\,n}<\dfrac16.$$

Checking odd values one by one:

For $$n=1$$, $$\bigl(\tfrac34\bigr)^1=\tfrac34=0.75>0.1667$$ (fails).

For $$n=3$$, $$\bigl(\tfrac34\bigr)^3=\dfrac{27}{64}\approx0.4219>0.1667$$ (fails).

For $$n=5$$, $$\bigl(\tfrac34\bigr)^5=\dfrac{243}{1024}\approx0.2373>0.1667$$ (fails).

For $$n=7$$, $$\bigl(\tfrac34\bigr)^7=\dfrac{2187}{16384}\approx0.1335<0.1667$$ (success).

Thus $$n=7$$ is the first (least) odd natural number for which $$\bigl(\tfrac34\bigr)^{\,n}<\dfrac16$$ and consequently $$B_n>A_n$$. For every odd $$n\ge7$$, the power $$\bigl(\tfrac34\bigr)^{\,n}$$ is even smaller, so the inequality continues to hold, and we already observed it always holds for even $$n$$.

Therefore the least odd natural number $$p$$ such that $$B_n>A_n$$ for all $$n\ge p$$ is

$$p=7.$$

Hence, the correct answer is Option B.