Question
In the expansion of $${\left( {x + \sqrt {{x^2} - 1} } \right)^6} + {\left( {x - \sqrt {{x^2} - 1} } \right)^6},$$ the number of terms is
A.
7
B.
14
C.
6
D.
4
Answer :
4
Solution :
On expansion and simplification,
expression $$ = 2\left\{ {^6{C_0}{x^6} + {\,^6}{C_2}{x^4}\left( {{x^2} - 1} \right) + {\,^6}{C_4}{x^2}{{\left( {{x^2} - 1} \right)}^2} + {\,^6}{C_6}{{\left( {{x^2} - 1} \right)}^3}} \right\}$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2\left\{ {\left( {^6{C_0} + {\,^6}{C_2} + {\,^6}{C_4} + {\,^6}{C_6}} \right){x^6} + \left( {{ - ^6}{C_2} - {\,^6}{C_4} \times 2 - {\,^6}{C_6} \times 3} \right){x^4} + \left( {^6{C_4} + {\,^6}{C_6} \times 3} \right){x^2} - {\,^6}{C_6}} \right\}.$$