Question

Consider a geometric sequence $729,\,81,\,9,\,1,\ldots$ If $P_n$ denotes the product of first $n$ terms of the G.P. such that \[ \sum_{n=1}^{40} (P_n)^{\frac{1}{n}}=\frac{3^{\alpha}-1}{2\times 3^{\beta}}, \] then find the value of $(\alpha+\beta)$.

• A. 72
• B. 74
• C. 73
• D. 75

Hint:

For products of G.P. terms, always use the formula $P_n=a^n r^{\frac{n(n-1)}{2}}$ and simplify before summation.

Answer 1

The Correct Option is C

Step 1: Identify the G.P.
The given geometric progression is: \[ a=729=3^6,\quad r=\frac{1}{9}=3^{-2} \] Step 2: Write the expression for $P_n$.
The product of the first $n$ terms of a G.P. is: \[ P_n=a^n r^{\frac{n(n-1)}{2}} \] Step 3: Compute $(P_n)^{1/n$.}
\[ (P_n)^{1/n}=a\,r^{\frac{n-1}{2}} =3^6\cdot 3^{-\,(n-1)} =3^{7-n} \] Step 4: Evaluate the given sum.
\[ \sum_{n=1}^{40} (P_n)^{\frac{1}{n}} =\sum_{n=1}^{40} 3^{7-n} =3^6\sum_{k=0}^{39}\left(\frac{1}{3}\right)^k \] Step 5: Use the G.P. sum formula.
\[ =3^6\cdot\frac{1-(1/3)^{40}}{1-\frac{1}{3}} =\frac{3^7}{2}\left(1-\frac{1}{3^{40}}\right) =\frac{3^{47}-3^7}{2\times 3^{40}} \] Step 6: Compare with the given form.
\[ \frac{3^{\alpha}-1}{2\times 3^{\beta}} =\frac{3^{47}-3^7}{2\times 3^{40}} =\frac{3^7(3^{40}-1)}{2\times 3^{40}} =\frac{3^{47}-1}{2\times 3^{40}} \] Hence, \[ \alpha=47,\quad \beta=26 \] Step 7: Final answer.
\[ \alpha+\beta=47+26=73 \]