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JEE Mains · Maths · STD 11 - 7. binomial theoram

\(\left(1+x^{ n }+x^{253}\right)^{10}\), ( जहाँ \(n \leq 22\) कोई धन पूर्णांक हैं) के प्रसार में \(x^{1012}\) का गुणांक हैं

  1. A \(1\)
  2. B \(^{10}{C_4}\)
  3. C \(4n\)
  4. D \(^{253}{C_4}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(^{10}{C_4}\)

Step-by-step Solution

Detailed explanation

Given expansion \(\left(1+x^{n}+x^{253}\right)^{10}\) Let \(x^{1012}=(1)^{a}\left(x^{n}\right)^{b} \cdot\left(x^{253}\right)^{c}\) Here \(a, b, c, n\) are all \(+ve\) integers and \(a\) \(\leq 10, b \leq 10, c \leq 4, n \leq 22, a+b+c=10\) Now \(b n+253 c=1012\)…
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