JEE Mains · Maths · STD 11 - 7. binomial theoram
If the second, third and fourth terms in the expansion of \((x+y)^{\mathrm{n}}\) are \(135\),\(30\) and \(\frac{10}{3}\), respectively, then \(6\left(n^3+x^2+y\right)\) is equal to .............
- A \(305\)
- B \(806\)
- C \(604\)
- D \(204\)
Answer & Solution
Correct Answer
(B) \(806\)
Step-by-step Solution
Detailed explanation
\( { }^n C_1 x^{n-1} y=135 \) ...........(\(i\)) \( { }^n C_2 x^{n-2} y^2=30 \) ............(\(ii\)) \( { }^n C_3 x^{n-3} y^3=\frac{10}{3} \) ............(\(iii\)) \( \text { By } \frac{(i)}{(i i)} \) \( \frac{{ }^n C_1}{{ }^n C_2} \frac{x}{y}=\frac{9}{2} \) ............(\(iv\))…
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