# 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. xn – 1. a1 + nC2. xn – 2. a2 + ……… + nCr. xn – r. ar + ……… + nCn. x0. an = $$\sum_{r=0}^{n}$$ nCr.xn – r. ar. and (x – a)nnC0. xn – (x)nnC1. xn – 1a + nC2 xn – 2. a2 …….. + ( – 1)r nCr. xn – r. ar + ……… + (- 1)n nCn.an → The expansion of (x + a)n contains (n + 1) terms.

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

→ In the expansion, (r + 1)th term is called the general term. It is denoted by Tr + 1.
∴ Tr + 1 = cCr. xn – r. ar, (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 pth 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 + ………. + Cn = 2n

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

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

→ $$\sum_{r=0}^{n}$$ nCr = 2n

→ $$\sum_{r=0}^{n}$$ r. nCr = n. 2n – 1

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

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

→ a. C0 + (a + d). C1 + (a + 2d). C2 + ……… + (a + nd). Cn = (2a + nd) 2n – 1

→ C0Cr + C1Cr + 1 + C2Cr + 2 + ………… + Cn – r. Cn = 2nCn + r

→ If f(x) = (a0 + a1x + a2x2 + ……… amxm)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 + ……… + ……… + $$\frac{n(n+1)(n+2) \ldots \ldots(n+r-1)}{r !}$$ xr + ……. to ∞

→ (1 + x)-n = 1,- nx + $$\frac{n(n+1)}{2 !}$$ x2 + …….. + $$\frac{(-1)^{r} n(n+1)(n+2) \ldots(n+r-1)}{r !}$$ xr + …….. ∞ → 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 = nCo xn + nC1 xn-1 a + nC2 xn-2 a2 + . … + nCr xn-rar + …… + nCnan

→ 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………….. nCn are called binomial coefficients and these are simply denoted by C0, C1, C2 …. CN.
nC0 = 1, nCN = 1, nC1 = n, nCr = nCn-r

→ In the expansion, (r + 1)th term is called the general term. It is denoted by
Tr+1. Thus Tr+1 = nCrxn-rar

→ (x + a)n = $$\sum_{r=0}^{n}$$nCrxn-rar

→ (x + a)n = $$\sum_{r=0}^{n}$$nCrxn-r(-a)r = $$\sum_{r=0}^{n}$$(-1)n nCrxn-r(-a)r = nC0xnnC1xn-1a + nC1xn-2a2 – ……….. + (-1)n nCn an

→ (1 + x)n = $$\sum_{r=0}^{n}$$nCrxr = nC0 + nC0x + ……….. + nCn xn = C0 + C1x + C2x2 + ………. + Cnxn

→ 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)rxr + …….
• (1 – x)-1 = 1 + x + x2 + x3 + … + xr + …….
• (1 + x)-2 = 1 – 2x + 3x2 – 4x3 + … + (-1)r (r + 1)xr + ………..
• (1 – x)-2 = 1 + 2x + 3x2 + 4x3 + … +(r + 1)xr + ….
• (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.xn-1a2 + …………… + nCr.xn-rar + ……….. + nCn.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 = kC0xk.a0 + kC1xk-2a2 + kC2.xk-2a2 + …+ kCr.xk-r.ar + ………….. + kCx0ak

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