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Binomial Theorem question

2020 · 4 Sep · Shift 1 · Q24
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  4. /Binomial Theorem
  5. /2020 · 4 Sep · Shift 1 · Q24

Binomial Theorem question

2020 · 4 Sep · Shift 1 · Q24

JEE MainMathematicsBinomial TheoremNumerical+4 / −1
Let (2x2+3x+4)10=∑r=020arxr{\left( {2{x^2} + 3x + 4} \right)^{10}} = \sum\limits_{r = 0}^{20} {{a_r}{x^r}}(2x2+3x+4)10=r=0∑20​ar​xr Then a7a13{{{a_7}} \over {{a_{13}}}}a13​a7​​ is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
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