Question

expand using binomial theorem (x^2 + 3y)^5​

Answer 1

Answer:

GIVEN EQUATION -

${( {x}^{2} + 3y) }^{5}$

USING BINOMIAL EQUATION -

${( {x}^{2} + 3y)}^{5} = \binom{5}{0} { ({x}^{2} )}^{5} {(3y)}^{0} + \binom{5}{1} { ({x}^{2} )}^{4} {(3y)}^{1} + \binom{5}{2} {( {x}^{2}) }^{3} {(3y)}^{2} + \binom{5}{3} { ({x}^{2}) }^{2} {(3y)}^{3} + \binom{5}{4} {( {x}^{2}) }^{1} {(3y)}^{4} + \binom{5}{5} {( {x}^{2}) }^{0} {(3y)}^{5}$

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