JEE Advanced · Mathematics · 6. Binomial Theorem
Let \(a_0, a_1, \ldots, a_{23}\) be real numbers such that
\(\left(1+\frac{2}{5} x\right)^{23}=\sum_{i=0}^{23} a_i x^i\)
for every real number \(x\). let \(a_r\) be the largest among the numbers \(a_j\) for \(0 \leq j \leq 23\).
The the value of \(r\) is ______
- A 2
- B 4
- C 6
- D 8
Answer & Solution
Correct Answer
(C) 6
Step-by-step Solution
Detailed explanation
For \(\mathrm{x}=1\)
\(\left(1+\frac{2}{5}\right)^{23}=a_0+a_1+a_2+\ldots+a_{23}\)
for numerically greatest term
\(\frac{n+1}{1+\left|\frac{a}{b}\right|}=\frac{23+1}{1+\frac{5}{2}}=\frac{48}{7}\)
\(\Rightarrow\left[\frac{48}{7}\right]=6=\mathrm{m}\) (where [.] greatest integer function)
so, \(\mathrm{T}_7\) is numerical greatest term
Hence \(r=6\)
\(\left(1+\frac{2}{5}\right)^{23}=a_0+a_1+a_2+\ldots+a_{23}\)
for numerically greatest term
\(\frac{n+1}{1+\left|\frac{a}{b}\right|}=\frac{23+1}{1+\frac{5}{2}}=\frac{48}{7}\)
\(\Rightarrow\left[\frac{48}{7}\right]=6=\mathrm{m}\) (where [.] greatest integer function)
so, \(\mathrm{T}_7\) is numerical greatest term
Hence \(r=6\)
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