TS Inter 2nd Year Maths 2A Binomial Theorem Formulas

Learning these TS Inter 2nd Year Maths 2A Formulas Chapter 6 Binomial Theorem will help students to solve mathematical problems quickly.

TS Inter 2nd Year Maths 2A Binomial Theorem Formulas

→ If a, x are real numbers and n is a positive integer, then

• (x + a)ⁿ = ⁿC₀ xⁿ. a⁰ + ⁿC₁. x^n-1 . a¹ + ⁿC₂. x^n-2. a² + ………… + ⁿC_r. x^n-r. a^r + ……………. + (-1)^{r n}C_n. x⁰ . aⁿ
• (x – a)ⁿ = ⁿC₀ xⁿ. a⁰ – ⁿC₁. x^n-1 . a¹ + ⁿC₂. x^n-2. a² + ………… + ⁿC_r. x^n-r. a^r + ……………. + (-1)^{r n}C_n. x⁰ . aⁿ

→ Each of the above two expansions contain (n + 1) terms in R.H.S.

→ The general term of (x + a)ⁿ is T_r+1 = ⁿC_r x^n-r . a^r (0 ≤ r ≤ n)

→ If n is fixed we write ⁿC_r = C_r and C₀, C₁ C₂ ………….. C_n are called the binomial coefficients.

• C₀ + C₁ + C₂ …………….. + C_n = 2ⁿ
• C₀ – C₁ + C₂ – C₃ + ……………. + (-1)ⁿC_n = 0
• C₀ + C₂ + C₄ + ………….. = C₁ + C₃ + C₅ + …………… = 2^n-1

→ \(\sum_{r=0}^n\)C_r = 2ⁿ
\(\sum_{r=1}^n\)r. C_r = n. 2^n-1
\(\sum_{r=2}^n\) r(r – 1). C_r = n(n – 1).2^n-2
\(\sum_{r=1}^n\) r² .C_r = n(n + 1) 2^n-2

→ (i) If n is even then the expansion of (x + a)ⁿ has only one middle term. It is
T_{\(\frac{n}{2}\)+1} = ⁿC_{\(\frac{n}{2}\)} . x^{\(\frac{n}{2}\)}. a^{\(\frac{n}{2}\)}

(ii) If n is odd, then the expansion of (x + a)ⁿ has two middle terms. They are
T_{\(\frac{n+1}{2}\)} = nC_{\(\left(\frac{n-1}{2}\right)\)} . x^{\(\frac{n+1}{2}\)}. a^{\(\frac{n-1}{2}\)} and T_{\(\frac{n+3}{2}\)} = nC_{\(\left(\frac{n+1}{2}\right)\)} . x^{\(\frac{n-1}{2}\)} . a^{\(\frac{n+1}{2}\)}

→ Numerically greatest term in the expansion of (1 + x)” :

• If \(\frac{(\mathrm{n}+1)|\mathrm{x}|}{1+|\mathrm{x}|}\) is not an integer and if its integral part \(\left[\frac{(n+1)|x|}{1+|x|}\right]\) = r, a positive integer then T_r+1 is the numerically greatest term in the expansion of (i + x)ⁿ.
• If \(\frac{(n+1)|x|}{1+|x|}\) is positive integer, say m, then |T_m| = |T_m+1| and hence T_m and T_m+1 both are numerically greatest terms in the expansion of (1 + x)ⁿ.

→ If x is a real number such that |x| < 1 and p. q are positive integers, then

→ If n is a positive integer and x is a real number such that j x j < 1, then

• If |x| is so small that x² and higher powers of x may be neglected, then (1 + x)ⁿ = 1 + nx.
• If |x| is so small that x³ and higher powers of x may be neglected, then
  (1 + x)ⁿ = 1 + nx + \(\frac{n(n-1)}{2 !}\)x².
• if |x| is so small that x⁴ and higher powers of x may be neglected, then
  (1 + x)ⁿ = 1 + nx + \(\frac{n(n-1)}{2 !}\)x² + \(\frac{n(n-1)(n-2)}{3 !}\)x³

→ (1 + x)^p/q = 1 + \(\frac{\frac{p}{q}}{1 !}\)x + \(\frac{\frac{p}{q}\left(\frac{p}{q}-1\right)}{2 !}\)x² + \(\frac{\frac{p}{q}\left(\frac{p}{q}-1\right)\left(\frac{p}{q}-2\right)}{3 !}\)x³ + ……. + \(\frac{\frac{p}{q}\left(\frac{p}{q}-1\right) \cdots\left(\frac{p}{q}-r+1\right)}{r !}\).x^r + …………
T_r+1 = \(\frac{\frac{p}{q}\left(\frac{p}{q}-1\right)\left(\frac{p}{q}-2\right) \ldots \ldots\left(\frac{p}{q}-r+1\right)}{r !}\). x^r for r ∈ N.

→ Let n ∈ N and a, b, c ∈ R, then (a + b + c)ⁿ contains \(\frac{(\mathrm{n}+1)(\mathrm{n}+2)}{2}\) terms.
Also (a +b + c)ⁿ = \(\sum_{{0 \leq p, q, r \leq n \\ p+q+r=n}} \frac{n !}{p ! q ! r !}\) a^p . b^q. c^r