If $a = \mathop \sum \limits_{n = 0}^\infty \frac{{{x^{3\;n}}}}{{\left( {3\;n} \right)\;!}},b = \mathop \sum \limits_{n = 1}^\infty \frac{{{x^{3\;n - 2}}}}{{\left( {3\;n - 2} \right)\;!}}$ and $c = \mathop \sum \limits_{n = 1}^\infty \frac{{{x^{3\;n - 1}}}}{{\left( {3n - 1} \right)\;!}},$ then the value of ${a^3} + {b^3} + {c^3} - 3\;abc$ is

JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

1 Hard

1 of 1

Analyzing +4 | -1

a = \sum \limits_{n = 0}^\infty \frac{{{x^{3\;n}}}}{{\left( {3\;n} \right)\;!}},b = \sum \limits_{n = 1}^\infty \frac{{{x^{3\;n - 2}}}}{{\left( {3\;n - 2} \right)\;!}}

and

c = \sum \limits_{n = 1}^\infty \frac{{{x^{3\;n - 1}}}}{{\left( {3n - 1} \right)\;!}},

then the value of

{a^3} + {b^3} + {c^3} - 3\;abc

View answer and explanation

Correct option: A

Option A (1) is correct because the expression

{a^3} + {b^3} + {c^3} - 3abc

can be factored, and using properties of complex numbers and exponential functions, it simplifies to 1.