1 + (3 + 1)(32 + 1)(34 + 1)(38 + 1)(316 + 1)(332 + 1) is equal to =?
A. $$\frac{{{3^{64}} - 1}}{2}$$
B. $$\frac{{{3^{64}} + 1}}{2}$$
C. 364 - 1
D. 364 + 1
Answer: Option B
Solution(By Examveda Team)
$$1 + \left( {3 + 1} \right)$$ $$\left( {{3^2} + 1} \right)$$ $$\left( {{3^4} + 1} \right)$$ $$\left( {{3^8} + 1} \right)$$ $$\left( {{3^{16}} + 1} \right)$$ $$\left( {{3^{32}} + 1} \right)$$$$ = 1 + \frac{1}{2}\left[ {\left( {3 - 1} \right)\left( {3 + 1} \right)\left( {{3^2} + 1} \right)\left( {{3^4} + 1} \right)\left( {{3^8} + 1} \right)\left( {{3^{16}} + 1} \right)\left( {{3^{32}} + 1} \right)} \right]$$
$$ = 1 + \frac{1}{2}\left[ {\left( {{3^2} - 1} \right)\left( {{3^2} + 1} \right)\left( {{3^4} + 1} \right)\left( {{3^8} + 1} \right)\left( {{3^{16}} + 1} \right)\left( {{3^{32}} + 1} \right)} \right]$$
$$ = 1 + \frac{1}{2}\left[ {\left( {{3^4} - 1} \right)\left( {{3^4} + 1} \right)\left( {{3^8} + 1} \right)\left( {{3^{16}} + 1} \right)\left( {{3^{32}} + 1} \right)} \right]$$
$$\eqalign{ & = 1 + \frac{1}{2}\left[ {\left( {{3^8} - 1} \right)\left( {{3^8} + 1} \right)\left( {{3^{16}} + 1} \right)\left( {{3^{32}} + 1} \right)} \right] \cr & = 1 + \frac{1}{2}\left[ {\left( {{3^{16}} - 1} \right)\left( {{3^{16}} + 1} \right)\left( {{3^{32}} + 1} \right)} \right] \cr & = 1 + \frac{1}{2}\left[ {\left( {{3^{32}} - 1} \right)\left( {{3^{32}} + 1} \right)} \right] \cr & = 1 + \frac{1}{2}\left[ {\left( {{3^{64}} + 1} \right)} \right] \cr & = \frac{{2 + {3^{64}} - 1}}{2} \cr & = \frac{{{3^{64}} + 1}}{2} \cr} $$
Related Questions on Surds and Indices
A. $$\frac{1}{2}$$
B. 1
C. 2
D. $$\frac{7}{2}$$
Given that 100.48 = x, 100.70 = y and xz = y2, then the value of z is close to:
A. 1.45
B. 1.88
C. 2.9
D. 3.7
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