# BINOMIAL THEOREM FOR POSITIVE INTEGRAL INDEX

If x and a are real, then  “n ϵ N i.e (x + a)n = nC₀xn + nC₁.xn-¹a+nC₂xm-².a²+ … +nCrxn-r.ar + … + nCn.x⁰an(x + a)n =$$\sum\limits_{r=0}^{n}{^{n}{{C}_{r}}{{x}^{n-r}}{{a}^{r}}}$$.

1. General term for (1 + x)nis  Tr + 1= nCr.xr

2. In the binomial expansion of (x + a)n, rthterm from end is (n – r + 2)th term from beginning

3. If n is odd then no .of terms in {(x + a)n + (x – a)n}, {(x + a)n – (x – a)n}. Have equal no. of terms = $$\left( \frac{n+1}{2} \right)\] {(x + a)n – (x – a)n} 4. If n is odd then {(x + a)n + (x – a)n} has [latex]\left( \frac{n}{2}+1 \right)$$ terms {(x + a)n – (x – a)n}  has   $$\left( \frac{n}{2} \right)$$ terms.

5. Middle term in a binomial expansion

If n is even

Then middle term of (x + a)n is $${{\left( \frac{n}{2}+1 \right)}^{th}}$$ term

6. If n is odd then Middle terms are $${{\left( \frac{n+1}{2} \right)}^{th}}$$ and $${{\left( \frac{n+3}{2} \right)}^{th}}$$ terms

7. STANDARD NOTATIONS

1. nC₀ + nC₁ + nC₂ + …+nCn = 2n
2. nC₀ + nC₂ + nC₄ + … = 2n-¹
3. nC₁+ nC₃ + nC₅ + … = 2n-¹
4. nC₀ + (a + d)nC₁ + (a + d)nC₂ + … + (a + nd) nCn = (2a + nd). 2n-¹
5. $$\sum\limits_{r=1}^{n}{r.{{\,}^{n}}{{C}_{r}}=n{{.2}^{n-1}}}$$.
6. $$\sum\limits_{r=2}^{n}{r{{(r-1)}^{n}}{{C}_{r}}=n(n-1){{2}^{n-2}}}$$.
7. $$\sum\limits_{r=2}^{n}{{{r}^{2}}{{.}^{n}}{{C}_{r}}=n(n+1){{2}^{n-2}}}$$.
8. nC₀ – nC₁ + nC₂ … + (-1)nCn = 0.
9. $$\frac{^{n}{{C}_{1}}}{^{n}{{C}_{0}}}+2.\frac{^{n}{{C}_{2}}}{^{n}{{C}_{1}}}+…+n\frac{^{n}{{C}_{n}}}{^{n}{{C}_{n-1}}}=\frac{n(n+1)}{2}$$.

8. Examples based on Integer part and fraction part

R=(5√5+ 11)²n + ¹

f = R – [R] [.] integer part

i.e [R] ≤ R

Then Rf = 4²n + ¹

9. (5 + 2√6)n = I + f where

I , n n ϵ N Then (I + f) ( I – f) = 1

Steps to solve these type problems:

1) Write the given expression =I +F

Where I is Integer, F is fractional part

2) Define G by replacing ‘+’ sign by ‘-‘.Note G always lies between 0 and 1

3) Either add or subtract G from expression in Step -I so that R.H.S is an integer.

4) G+F = 1 i.e. if G is added G = 1-F

i.e if G is subtracted then G-F = 0 = G =F

5) Obtain the value of desired expression given.

Greatest term: Let n ϵ N   and x ϵ R – {0} and $$m=\left[ \frac{(n+1)\,|x|}{1+|x|} \right]\,\,\,[.]$$ is integral part then

If $$\frac{(n+1)\,|x|}{1+|x|}$$ is not a integer, Then Tm+₁ is numerically greatest term in the binomial expansion of (1 + x)n

If $$\frac{(n+1)\,|x|}{1+|x|}$$ is an integer then Tmand Tm+₁ are numerically greatest terms in expansion (1 + x)n

Largest binomial coefficient: The largest among nC₀, nC₁, … nCn is (are):

If n is event integer then $$^{n}{{C}_{\left( \frac{n}{2} \right)}}$$ is largest.

If n is odd integer $$^{n}{{C}_{\left( \frac{n-1}{2} \right)}},{{\,}^{n}}{{C}_{\left( \frac{n+1}{2} \right)}}\,\,\,$$.

a) $${{[}^{n}}{{C}_{0}}+\frac{^{n}{{C}_{1}}}{2}.x+{{.}^{n}}{{C}_{2}}.\frac{{{x}^{2}}}{3}+…{{+}^{n}}{{C}_{n}}\frac{{{x}^{n}}}{n+1}=\frac{{{(1+x)}^{n+1}}-1}{(n+1)x}$$.

b) $${{{{(}^{n}}{{C}_{0}})}^{2}}-{{{{(}^{n}}{{C}_{1}})}^{2}}+{{{{(}^{n}}{{C}_{1}})}^{2}}+…+{{(-1)}^{n}}.{{{{(}^{n}}{{C}_{n}})}^{2}}=\{{{(-1)}^{nn}}{{C}_{n/2}}\,if\,n\,is\,odd$$.

c) $$-\frac{^{n}{{C}_{1}}}{2}+\frac{^{n}{{C}_{2}}}{3}+…+{{(-1)}^{n}}\frac{^{n}{{C}_{n}}}{n+1}=\frac{1}{n+1}$$.

MULTINOMINAL THEORM:

(x₁ + x₂ + … + xn)n = $$\sum\limits_{{{r}_{1}}+{{r}_{2}}+……+{{r}_{n}}=n}{\frac{n!}{{{r}_{1}}!.{{r}_{2}}!…{{r}_{k}}!}{{x}_{1.}}^{{{r}_{1}}}}{{x}_{2}}^{{{r}_{2}}}…{{x}_{k}}^{r}$$.

The general term in the above expression is = $$\frac{n!}{{{r}_{1}}!.{{r}_{2}}!…{{r}_{k}}!}{{x}_{1.}}^{{{r}_{1}}}{{x}_{2}}^{{{r}_{2}}}…\,{{x}_{k}}^{r}$$.

The no. of terms in expansion is n + k – 1Ck – 1

Particular cases:

i) (x + y + z)n = $$\sum\limits_{r+s+t=n}{\frac{n!}{r!.s!t!}}{{x}^{r}}{{y}^{s}}{{z}^{t}}$$.

No. of terms = n + ² C₂ terms

ii) (x + y + z + t)n = $$\sum\limits_{r+s+t=n}{\frac{n!}{r!.s!t!}}{{x}^{r}}{{y}^{s}}{{z}^{t}}$$.

Total terms = n + ³C₃

The greatest coefficient in the expansion of (x₁ + x₂ + … + xm)n] is $$\left[ \frac{n!}{{{(q!)}^{m-r}}{{[(q+1)!]}^{r}}} \right]$$.

Where q and r are Quotient and remainder respectively when n is divided by M

nCr = . n – ¹Cr – ₁ b

nCr + nCr – ₁ = n + ¹Cr

iii) $$\frac{^{n}{{C}_{r}}}{^{n}{{C}_{r-1}}}=\frac{n-r+1}{r}$$.

nCr = nCs then r = s or r + s = n

1. $$\sum\limits_{0\le r}{\sum\limits_{s\le n}{(r+s){{.}^{n}}{{C}_{r}}^{n}{{C}_{s}}=n.[{{2}^{2n+1}}{{-}^{2n-1}}{{C}_{n-1}}]}}$$.
2. $$\sum\limits_{0\le r\,\,<}{\sum\limits_{s\le n}{(r+s)\,{{(}^{n}}{{C}_{r}}^{n}{{C}_{s}})={{n}^{2}}{{.2}^{n}}}}$$.
3. If (1 + x + x²)n = $$\sum\limits_{r=0}^{2n}{{{a}_{r}}{{x}^{r}}}$$.

Then (1 + x + x²)n = \sum\limits_{r=0}^{n}{{{(-1)}^{r}}{{a}_{r}}^{n}{{C}_{r}}}=\left\{ \begin{align}& 0,\,if\,\,n\,\,is\,\,not\,a\,multiple\,\,of\,\,3 \\& ^{n}{{C}_{n/3}}\,if\,n\,is\,multiple\,of\,\,3 \\\end{align} \right.

No three consecutive binomial coefficients can be in (i) G.P and (ii) H.P If is odd then $$\sum\limits_{r=0}^{n}{{{(-1)}^{r}}\frac{1}{^{n}{{C}_{r}}}}={{0}^{n}}$$.

To find the remainder of using Binomial theorem

Find the remainder

Ex = 32³² divided by 7

(32)³² = (2⁵)³² = 2¹⁶⁰ = (3-1)¹⁶⁰[¹⁶⁰C₀3¹⁶⁰ + … + 1¹] = 3m + 1

32³² = 32³m + ¹

= 2¹⁵m + ⁵ = (2³)³m + ¹ .4

= 8⁵m + ¹ .4

= (7 + 1)⁵m + ¹ .4

= (⁵m + ¹ C₀.7⁵m + ¹ + … + 1) 4

Remainder = 4

mCr + mCr – nC1 + … + mCr = m + nCr

Where r < m, r < n m, n, r are z+