Binomial Theorem — Flashcards

Answer

Using the binomial expansion: $$ (x + 2)^4 = \sum_{r=0}^{4} {}^4C_r x^{4-r} (2)^r $$ Step-by-step: - $ r = 0 $: $ {}^4C_0 x^4 (2)^0 = 1 \cdot x^4 \cdot 1 = x^4 $ - $ r = 1 $: $ {}^4C_1 x^3 (2)^1 =…

Answer

The general term is: $$ T_{r+1} = {}^nC_r a^{n-r} b^r $$ Here, $ n = 7 $, $ a = 2x $, $ b = -3 $. We need the 5th term → $ r + 1 = 5 $ → $ r = 4 $ $$ T_5 = {}^7C_4 (2x)^{7-4} (-3)^4 = {}^7C_4 (2x)^…

Answer

Since $ n = 6 $ (even), there is **one** middle term: the $ \left(\frac{6}{2} + 1\right) = 4^{th} $ term. So, $ r + 1 = 4 $ → $ r = 3 $ $$ T_4 = {}^6C_3 (3x)^{6-3} (y)^3 = 20 \cdot (27x^3) \cdot y^3…

Answer

General term: $ T_{r+1} = {}^8C_r (1)^{8-r} (2x)^r = {}^8C_r \cdot 2^r \cdot x^r $ We want $ x^r = x^3 $ → $ r = 3 $ So, coefficient = $ {}^8C_3 \cdot 2^3 = 56 \cdot 8 = 448 $ Answer: 448…

Answer

Using binomial theorem for rational index: $$ (1 + x)^{-2} = 1 + (-2)x + \frac{(-2)(-3)}{2!}x^2 + \frac{(-2)(-3)(-4)}{3!}x^3 + \dots $$ Simplify: - $ = 1 - 2x + \frac{6}{2}x^2 - \frac{24}{6}x^3 + \…

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Total terms = $ n + 1 = 11 $. So, 4th term from end = $ (11 - 4 + 1) = 8^{th} $ term from start. So, $ r + 1 = 8 $ → $ r = 7 $ $$ T_8 = {}^{10}C_7 a^{10-7} (-b)^7 = 120 \cdot a^3 \cdot (-b^7) = -120…

Answer

Use this formula to: - Find a specific term (like 5th term) - Find the coefficient of a particular power - Determine if a term is independent of $ x $ - Locate middle or extreme terms Example: To fin…

Answer

The binomial expansion is: $$ (1 + x)^n = 1 + {}^nC_1 x + {}^nC_2 x^2 + \dots + {}^nC_n x^n $$ Or: $$ (1 + x)^n = \sum_{r=0}^{n} {}^nC_r x^r $$ Where $ {}^nC_r = \frac{n!}{r!(n-r)!} $ Example: $ …