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Binomial Theorem Part 3

Section: Mathematics

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Question : 34 of 53

Marks: +1, -0

Let the coefficients of three consecutive terms

T_r, T_{r+1}

T_{r+2}

in the binomial expansion of

(a+b)^{12}

be in a G.P. and let

p

be the number of all possible values of

r

q

be the sum of all rational terms in the binomial expansion

(\;\sqrt[4]{3}+\;\sqrt[3]{4})^{12}

p+q

[28 Jan 2025 Shift 2]

287
295
299
283

Solution: 👈: Video Solution

Coefficient of

\;T_r, T_{r+1}, T_{r+2} \longrightarrow \text{GP}

\;\Rightarrow (\binom{12}{r})^2 = \binom{12}{r-1} \cdot \binom{12}{r+1}

but no three consecutive binomial coefficient are in GP

\Rightarrow P=0

Now for

\left(3^{\frac{1}{4}}+4^{\frac{1}{3}}\right)^{12}, T_{r+1} = \binom{12}{r} (4)^{\frac{K}{3}} (3)^{\;\frac{12-K}{4}}

for rational terms

K =0,12

sum of rational terms

\;= \binom{12}{0} 4^0 \cdot 3^3 + \binom{12}{12} \cdot 4^4 \cdot 3^0

\;=27+256=283=q

\;\therefore p+q=283

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