If $$\left( {1 + x} \right)\left( {1 + {x^2}} \right)\left( {1 + {x^4}} \right).....\left( {1 + {x^{128}}} \right) = \sum\limits_{r = 0}^n {{x^r}} $$ then $$n$$ is
A.
255
B.
127
C.
63
D.
none of these
Answer :
255
Solution :
$$n$$ = index of the highest power of $$x = 1 + 2 + 4 + . . . . . + 128.$$
Releted MCQ Question on Algebra >> Sequences and Series
Releted Question 1
If $$x, y$$ and $$z$$ are $${p^{{\text{th}}}},{q^{{\text{th}}}}\,{\text{and }}{r^{{\text{th}}}}$$ terms respectively of an A.P. and also of a G.P., then $${x^{y - z}}{y^{z - x}}{z^{x - y}}$$ is equal to:
If $$a, b, c$$ are in G.P., then the equations $$a{x^2} + 2bx + c = 0$$ and $$d{x^2} + 2ex + f = 0$$ have a common root if $$\frac{d}{a},\frac{e}{b},\frac{f}{c}$$ are in-