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JEE Class main Answered

plz solve it
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Asked by swayamagarwal2114 | 18 Jul, 2022, 03:06: PM
answered-by-expert Expert Answer
e to the power of x cubed end exponent equals sum from n equals 0 to infinity of fraction numerator x to the power of 3 n end exponent over denominator n factorial end fraction space space a n d space space e to the power of negative x cubed end exponent equals sum from n equals 0 to infinity of fraction numerator left parenthesis negative 1 right parenthesis to the power of 3 n end exponent space x to the power of 3 n end exponent over denominator n factorial end fraction
rightwards double arrow integral subscript negative 1 end subscript superscript 1 left parenthesis e to the power of x cubed end exponent plus e to the power of negative x cubed end exponent right parenthesis space d x space equals integral subscript negative 1 end subscript superscript 1 sum from n equals 0 to infinity of open parentheses fraction numerator x to the power of 3 n end exponent over denominator n factorial end fraction plus fraction numerator left parenthesis negative 1 right parenthesis to the power of 3 n end exponent space x to the power of 3 n end exponent over denominator n factorial end fraction close parentheses
Answered by Renu Varma | 26 Jul, 2022, 06:20: PM
JEE main - Maths
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how
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