Inter 2nd Year Maths 2A Binomial Theorem Formulas

Use these Inter 2nd Year Maths 2A Formulas PDF Chapter 6 Binomial Theorem to solve questions creatively.

Intermediate 2nd Year Maths 2A Binomial Theorem Formulas

→ Let n be a positive integer and x, a be real numbers then
(x + a)ⁿ = ⁿC₀. xⁿ. a⁰ + ⁿC₁. x^{n – 1}. a¹ + ⁿC₂. x^{n – 2}. a² + ……… + ⁿC_r. x^{n – r}. a^r + ……… + ⁿC_n. x⁰. aⁿ = \(\sum_{r=0}^{n}\) ⁿC_r.x^{n – r}. a^r. and (x – a)ⁿ – ⁿC₀. xⁿ – (x)ⁿ – ⁿC₁. x^{n – 1}a + ⁿC₂ x^{n – 2}. a² …….. + ( – 1)^{r n}C_r. x^{n – r}. a^r + ……… + (- 1)^{n n}C_n.aⁿ

→ The expansion of (x + a)ⁿ contains (n + 1) terms.

→ In the expansion, the coefficients ⁿC₀, ⁿC₁, ⁿC₂, ….. ⁿC_n are called binomial coefficients and these are simply denoted by C₀, C₁, C₂, …….. C_n,.

→ In the expansion, (r + 1)^th term is called the general term. It is denoted by T_{r + 1}.
∴ T_{r + 1} = ^cC_r. x^{n – r}. a^r, (0 ≤ r ≤ n)

→ The number of terms in the expansion of (a + b + c)ⁿ = \(\frac{(n+1)(n+2)}{2}\)

→ If n is even in the expansion of (x + a)ⁿ, the middle term = T_{\(\left(\frac{n}{2}+1\right)\)}

→ If n is odd in the expansion of (x + a)ⁿ, it has two middle terms which are T_{\(\left(\frac{n+1}{2}\right)\)}, T_{\(\left(\frac{n+3}{2}\right)\)}.

→ If \(\frac{(n+1)|x|}{|x|+1}\) = p, a positive integer then p^th and (p + 1)^th terms are the numerically greatest terms in the expansion of (1 + x)ⁿ.

→ If \(\frac{(n+1)|x|}{|x|+1}\) = P + F where p is a positive integer and 0 < F < 1 then (p + 1)^th the numerically greatest term in the expansion of (1 + x)ⁿ

→ C₀ + C₁ + C₂ + ………. + C_n = 2ⁿ

→ C₀ – C₁ + C₂ – C₃ + ……… + (- 1)ⁿC_n = 0

→ C₀ + C₂ + C ₄ + …………… = C₁ + C₃ + C₅ + …………….. = 2^{n – 1}

→ \(\sum_{r=0}^{n}\) ⁿC_r = 2ⁿ

→ \(\sum_{r=0}^{n}\) r. ⁿC_r = n. 2^{n – 1}

→ \(\sum_{r=2}^{n}\) r(r – 1). ⁿC_r = n(n – 1). 2^{n – 2}

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

→ a. C₀ + (a + d). C₁ + (a + 2d). C₂ + ……… + (a + nd). C_n = (2a + nd) 2^{n – 1}

→ C₀C_r + C₁C_{r + 1} + C₂C_{r + 2} + ………… + C_{n – r}. C_n = ²ⁿC_{n + r}

→ If f(x) = (a₀ + a₁x + a₂x² + ……… a_mx^m)ⁿ then

• Sum of the coefficients = f(1)
• Sum of the coefficients of even powers of x is \(\frac{f(1)+f(-1)}{2}\)
• Sum of the coefficients of odd powers of x is \(\frac{f(1)-f(-1)}{2}\)

→ Let n be a positive integer and x is a ,real number such that |x| < 1 then

→ (1 – x)^-n = 1 + nx + \(\frac{n(n+1)}{2 !}\) x² + \(\frac{n(n+1)(n+2)}{3 !}\) x² + ……… + ……… + \(\frac{n(n+1)(n+2) \ldots \ldots(n+r-1)}{r !}\) x^r + ……. to ∞

→ (1 + x)^-n = 1,- nx + \(\frac{n(n+1)}{2 !}\) x² + …….. + \(\frac{(-1)^{r} n(n+1)(n+2) \ldots(n+r-1)}{r !}\) x^r + …….. ∞

→ If |x| < 1, then for p, q ∈ N

→ (1 – x)^-p/q = 1 + \(\frac{p}{1 !}\left(\frac{x}{q}\right)\) + \(\frac{p(p+q)}{2 !}\left(\frac{x}{q}\right)^{2}\) + ………. + \(\frac{p(p+q) \ldots \ldots(p+(r-1) q)}{r !}\) \(\left(\frac{x}{q}\right)^{r}\) + …….. ∞

→ (1 + x)^-p/q = 1 – \(\frac{p}{1 !}\left(\frac{x}{q}\right)\) + \(\frac{p(p+q)}{2 !}\left(\frac{x}{q}\right)^{2}\) + ………. + \(\frac{(-1)^{r} p(p+q) \ldots(p+(r-1) q)}{r !}\) \(\left(\frac{x}{q}\right)^{r}\) + …….. ∞

→ (1 + x)^-p/q = 1 + \(\frac{p}{1 !}\left(\frac{x}{q}\right)\) + \(\frac{(p)(p-q)}{1 .2}\left(\frac{x}{q}\right)^{2}\) + …………. + \(\frac{(p)(p-q)(p-2 q) \ldots \ldots .[p-(r-1) q]}{(r) !}\) \(\left(\frac{x}{\cdot q}\right)^{r}\) + ……… ∞

→ (1 – x)^-p/q = 1 – \(\frac{p}{1 !}\left(\frac{x}{q}\right)\) + \(\frac{(p)(p-q)}{1.2}\left(\frac{x}{q}\right)^{2}\) – …………. + (- 1)^r \(\) \(\left(\frac{x}{q}\right)^{r}\) + ……… ∞

Binomial Theorem for integral index:
If n is a positive integer then (x + a)ⁿ = ⁿC_o xⁿ + ⁿC₁ x^n-1 a + ⁿC₂ x^n-2 a² + . … + ⁿC_r x^n-ra^r + …… + ⁿC_naⁿ

→ The expansion of (x + a)ⁿ contains (n + 1) terms.

→ In the expansion, the sum of the powers of x and a in each term is equal to n.

→ In the expansion, the coefficients ⁿC₀, ⁿC₁. ⁿC₂………….. ⁿC_n are called binomial coefficients and these are simply denoted by C₀, C₁, C₂ …. C_N.
ⁿC₀ = 1, ⁿC_N = 1, ⁿC₁ = n, ⁿC_r = ⁿC_n-r

→ In the expansion, (r + 1)^th term is called the general term. It is denoted by
T_r+1. Thus T_r+1 = ⁿC_rx_n-ra_r

→ (x + a)ⁿ = \(\sum_{r=0}^{n}\)ⁿC_rx_n-ra_r

→ (x + a)ⁿ = \(\sum_{r=0}^{n}\)ⁿC_rx_n-r(-a)_r = \(\sum_{r=0}^{n}\)(-1)^{n n}C_rx_n-r(-a)_r = ⁿC₀xⁿ – ⁿC₁x^n-1a + ⁿC₁x^n-2a² – ……….. + (-1)^{n n}C_n aⁿ

→ (1 + x)ⁿ = \(\sum_{r=0}^{n}\)ⁿC_rx_r = ⁿC₀ + ⁿC₀x + ……….. + ⁿC_n xⁿ = C₀ + C₁x + C₂x² + ………. + C_nxⁿ

→ Middle term(s) in the expansion of (x + a)ⁿ.

• If n is even, then (\(\frac{n}{2}\) + 1)th term is the middle term
• If n is odd, then \(\frac{n+1}{2}\) th and \(\frac{n+3}{2}\) th terms are the middle terms.

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

• If \(\frac{(n+1)|x|}{|x|+1}\) = p, a integer then plu1 and (p + 1) th terms are the numerically greatest terms in the expansion of (1 + x).
• If \(\frac{(n+1)|x|}{|x|+1}\) = p + F where pis a positive integer and 0< F < 1 then (p+1) th term is the numerically greatest term in the expansion of (1 + x).

→ Binomial Theorem for rational index: If n is a rational number and
|x| < 1, then 1 + nx + \(\frac{n(n-1)}{2 !}\) x² + \(\frac{n(n-1)(n-2)}{3 !}\)x³ + ………… = (1 + x)ⁿ

→ If |x| < 1 then

• (1 + x)^-1= 1 – x + x² – x³ + … + (-1)^rx^r + …….
• (1 – x)^-1 = 1 + x + x² + x³ + … + x^r + …….
• (1 + x)^-2 = 1 – 2x + 3x² – 4x³ + … + (-1)^r (r + 1)x^r + ………..
• (1 – x)^-2 = 1 + 2x + 3x² + 4x³ + … +(r + 1)x^r + ….
• (1 – x)^-n = 1 – nx + \(\frac{n(n-1)}{2 !}\) x² – \(\frac{n(n-1)(n-2)}{3 !}\)x³ + …………..
• (1 – x)^-n = 1 + nx + \(\frac{n(n-1)}{2 !}\) x² + \(\frac{n(n-1)(n-2)}{3 !}\)x³ + ………

→ If |x| < 1 an dn is a positive integer, then

• (1 – x)^-n = 1 + ⁿC₁x + ⁽ⁿ⁺¹⁾C₂x² + ⁽ⁿ⁺²⁾C₃x³ + ………….
• (1 + x)^-n = 1 – ⁿC₁x + ⁽ⁿ⁺¹⁾C₂x² – ⁽ⁿ⁺²⁾C₃x³ + ………….

→ When |x| < 1
(1 – x)^-p/q = 1 + \(\frac{p}{1 !}\left(\frac{x}{q}\right)+\frac{p(p+q)}{2 !}\left(\frac{x}{q}\right)^{2}+\frac{p(p+q)(p+2 q)}{3 !}\left(\frac{x}{q}\right)^{3}\) + …………………∞

→ When |x| < 1
(1 + x)^-p/q = 1 – \(\frac{p}{1 !}\left(\frac{x}{q}\right)_{+} \frac{p(p+q)}{2 !}\left(\frac{x}{q}\right)^{2} \quad \frac{p(p+q)(p+2 q)}{3 !}\left(\frac{x}{q}\right)^{3}\) + …………………∞

Binomial Theorem:
Let n be a positive integer and x, a be real numbers, then (x + a)ⁿ = ⁿC₀.xⁿa° + ⁿC₁.xⁿa¹ + ⁿC₂.x^n-1a² + …………… + ⁿC_r.x^n-ra^r + ……….. + ⁿC_n.x⁰aⁿ
Proof :
We prove this theorem by u sing the principle of mathematical induction (on n).
When n = 1, (x + a)¹ =(x + a)¹ = x + a = ¹C₀x¹a°+ ¹C₁x°a¹
Thus the theorem is true for n = 1
Assume that the theorem is true for n = k ≥ 1 (where k is a positive integer). That is
(x+a)^k = ^kC₀x^k.a⁰ + ^kC₁x^k-2a² + ^kC₂.x^k-2a² + …+ ^kC_r.x^k-r.a^r + ………….. + ^kCx⁰a^k

Now we prove that the theorem is true when n = k + 1 also
(x + a)^k+1 = (x + a)(x + a)^k

Therefore the theorem is true for n = k + 1
Hence, by mathematical induction, it follows that the theorem is true of all positive integer n