Binomial theorem — expansion, general term: Question bank

2.7.12 · D5 · Maths › Statistics & Probability — Intermediate › Binomial theorem — expansion, general term

Jo formula puri jagah attack mein hai — jahan $n$ ek non-negative integer hai ( $n = 0, 1, 2, \dots$ ) jab tak koi line aur na kahe: $(a + b)^{n} = \sum_{r = 0}^{n} (r n) a^{n - r} b^{r}, T_{r + 1} = (r n) a^{n - r} b^{r} .$

Yaad karo (dekho Combinations nCr): $(r n)$ $b$ donate karne ke liye $r$ brackets choose karne ke tarike ki sankhya hai — woh picture dimaag mein rakho; neeche ke zyaadatar traps usi par wapas jaane se dissolve ho jaate hain.

True or false — justify

Kisi term ka coefficient hamesha

(r n)

hota hai

False — yeh tab hi sach hai jab

a

aur

b

mein koi numeric ya power factors na hon (jaise

(1 + x)^{n}

mein). Generally kisi term ka coefficient hota hai binomial coefficient

(r n)

times

a^{n - r}

aur

b^{r}

ke numeric parts.

(a + b)^{n}

aur

(b + a)^{n}

ka expansion same hota hai

True as a whole, lekin term-by-term ye reverse order mein hote hain:

a

aur

b

swap karne se

r \to n - r

ho jaata hai, aur

(r n) = (n - r n)

collection ko identical rakhta hai.

(a - b)^{n}

ke saare coefficients positive hote hain

False —

b \to - b

likhne par har term mein

(- 1)^{r}

aa jaata hai, toh terms ke coefficients alternate karte hain:

+, -, +, -, \dots

jaise

r

badhta hai.

(a + b)^{n}

mein collect karne ke baad hamesha exactly

n + 1

terms hote hain

True non-negative integer

n

ke liye; har value

r = 0, 1, \dots, n

ek distinct power pairing deta hai, aur koi aapas mein collapse nahin hote.

Expansion mein

a

aur

b

dono ki powers badhti hain

False — exponents ka sum

n

hona chahiye, toh jaise

b

ki power

r

badhti hai,

a

ki power

n - r

ghatni padti hai. Sirf ek hi badh sakti hai.

(r n) = (n - r n)

, toh binomial coefficients ki row symmetric hoti hai

True — yeh choose karna ki

r

brackets

b

dein, yahi hai

n - r

brackets

a

dein; binomial-coefficient list aage-peeche same padhti hai.

(x + 2)^{4}

mein middle binomial coefficient

(2 4) = 6

hai

True as stated — binomial coefficient

6

hai. Lekin term ka actual coefficient

6 \cdot 2^{2} = 24

hai, kyunki

b = 2

ki wajah se

2^{r}

aata hai. Trap dono words ko confuse karna hai.

(1 + x)^{n}

x = 1

par evaluate karne se saare term coefficients ka sum milta hai

True —

x = 1

substitute karne se har power

1

ban jaati hai, aur

\sum (r n) = 2^{n}

milta hai, jo precisely coefficient sum hai.

Term

T_{r + 1}

expansion ka

r

-va term hai

False — yeh

(r + 1)

-va term

T_{r + 1}

hai. Counter

r

zero se shuru hota hai, toh yeh hamesha term ki position number se ek peeche rehta hai.

Spot the error

Student

(2 x - \frac{3}{x ^{2}})^{6}

ka general term

(r 6) (2 x)^{6 - r} (\frac{3}{x ^{2}})^{r}

likhta hai

Error: unhone minus sign drop kar diya. Poora signed quantity

- \frac{3}{x ^{2}}

b

hai, toh

(- 3)^{r}

hona chahiye, jo term ka sign

r

ke saath alternate karta hai.

Student

(x + 2)^{4}

x^{4} + 4 x^{3} + 6 x^{2} + 4 x + 1

expand karta hai

Error: unhone sirf binomial coefficients

(r 4)

use kiye aur

b = 2

ko bhool gaye. Numeric factor

2^{r}

include karne par milta hai

x^{4} + 8 x^{3} + 24 x^{2} + 32 x + 16

Student kehta hai

(a + b)^{n} = (r n) a^{n - r} b^{r}

r

-ve term" ke liye aur

r

directly padh leta hai

Error: yeh line ek single term

T_{r + 1}

hai, poora sum nahin, aur iska position

r + 1

hai. Position ko

r

padhne se har jawab ek se shift ho jaata hai.

Student

(x^{2} + \frac{1}{x})^{9}

mein

x^{6}

term ke liye

18 - 2 r = 6

set karke solve karta hai

Error: unhone

(\frac{1}{x})^{r}

se aane wala

x^{- r}

bhool gaya. Sahi exponent hai

18 - 2 r - r = 18 - 3 r

; use

6

set karne par

r = 4

milta hai.

Student claim karta hai

(3 7) = \frac{7 !}{3 !}

Error: denominator mein

(n - r)!

missing hai. Sahi se

(3 7) = \frac{7 !}{3 ! 4 !} = 35

; dekho Factorials.

Student

(1 + x)^{7}

ka middle term sirf

T_{4}

bolta hai

Error:

n = 7

odd hai, toh do middle terms hain,

T_{4}

aur

T_{5}

(

r = 3

aur

r = 4

se). Ek single middle term tab hi hota hai jab

n

even ho.

Student kehta hai

(1 + 2 x)^{5}

mein

x^{3}

ka coefficient

(3 5) = 10

hai

Error:

b = 2 x

ki wajah se

2^{3}

contribute hota hai, toh term ka coefficient

(3 5) \cdot 2^{3} = 10 \cdot 8 = 80

hai,

10

nahin. Binomial coefficient

10

hai; coefficient

80

hai.

Why questions

(r n)

counting number aata hi kyun hai, koi aur formula kyun nahin

Kyunki har raw product

a^{n - r} b^{r}

har us choice ke liye ek baar aata hai ki

r

brackets

b

dein (upar ka box figure dekho); aisi choices ki sankhya

(r n)

hai, toh identical products us count tak add hote hain. Dekho Permutations vs Combinations ki order kyun ignore hoti hai.

Binomial coefficient ek combination

(r n)

kyun hai, permutation kyun nahin

Kyunki multiply hone ke baad chosen

r

brackets indistinguishable hain — product

b \cdot b \dots b

ko parwah nahin ki humne unhein kis order mein pick kiya, toh hum sets count karte hain, sequences nahin.

x^{k}

wala term" solve karne se pehle exponents ko

x

ki single power mein combine kyun karte hain

Kyunki sirf ek clean quantity —

x

ka total exponent — ko target

k

se set kiya ja sakta hai. Minimally dikhate hain:

(x^{2} + \frac{1}{x})^{9}

mein ek term hai

(x^{2})^{9 - r} (x^{- 1})^{r} = x^{18 - 2 r} \cdot x^{- r} = x^{18 - 3 r}

; ab ek equation

18 - 3 r = k

se solve hota hai.

x^{18 - 2 r} x^{- r}

alag chhod do toh

k

se set karne ke liye kuch single nahin hai.

Constant (independent-of-

x

) term exponent ko

0

set karne se kyun aata hai

Kyunki

x^{0} = 1

; zero net power wala term

x

bilkul carry nahin karta — jaise

x^{18 - 3 r}

constant hai jab

18 - 3 r = 0

, yaani

r = 6

— aur yahi "

x

se independent" ka matlab hai.

Negative signs ko power

(- 3)^{r}

ke andar rakhhna zaroori kyun hai, unhe ek baar bahar factor out karne ki jagah

Kyunki sign

r

ke har increment ke saath flip hota hai:

(- 3)^{r}

even

r

ke liye

+

hai aur odd

r

ke liye

-

(jaise

r = 2 \to + 9

r = 3 \to - 27

), toh ek sign aage nikaalna aadhe terms ke liye galat hoga.

Binomial coefficients Pascal's Triangle ki ek row se kyun match karte hain

Kyunki Pascal's rule

(r n) = (r - 1 n - 1) + (r n - 1)

(a + b)^{n} = (a + b) (a + b)^{n - 1}

expand karne ko mirror karta hai — har term ek bracket se inherit karta hai jo

a

b

choose karta hai.

Saare binomial coefficients ka sum

2^{n}

kyun hota hai

Kyunki

n

brackets mein se har ek independently

a

b

choose karta hai — har ek ke liye do options, kul

2^{n}

selections — aur har selection exactly ek coefficient mein land karta hai. Ise dekhne ke liye

a = b = 1

set karo.

Probability mein same

(r n) p^{r} q^{n - r}

shape kyun aata hai

Kyunki

n

independent trials ki ek run "success" ko kuch

r

trials mein exactly

(r n)

tarike se choose karti hai, jo bracket-picking count se match karta hai; dekho Binomial Probability Distribution.

Edge cases

(a + b)^{0}

kya hai aur kya formula survive karta hai

Yeh

1

ke barabar hai: sum mein ek hi term hai

r = 0

(0 0) a^{0} b^{0} = 1

. "

n + 1 = 1

term" rule hold karta hai, aur

0! = 1

(0 0) = 1

well-defined rehta hai.

Jab

b = 0

ho toh expansion ka kya hoga

r \geq 1

wala har term

0^{r} = 0

carry karta hai aur vanish ho jaata hai, sirf

r = 0

bachta hai:

(0 n) a^{n} = a^{n}

. Theorem sahi se

(a + 0)^{n} = a^{n}

par reduce ho jaata hai.

Jab

a = 0

ho toh expansion ka kya hoga (mirror case)

Isi reasoning se,

n - r \geq 1

wala har term

0^{n - r} = 0

carry karta hai aur vanish ho jaata hai, sirf

r = n

bachta hai:

(n n) b^{n} = b^{n}

. Toh

(0 + b)^{n} = b^{n}

, bilkul

b = 0

case ka mirror.

Kya

(r n)

tab bhi sense karta hai jab

r = 0

r = n

Haan — dono

1

ke barabar hain, kyunki

b

ke liye koi bracket na choose karne ya saare brackets choose karne ka exactly ek hi tarika hai. Ye row ke do end coefficients hain.

Kya general term

r = n

(last term) ke liye valid hai

Haan —

T_{n + 1} = (n n) a^{0} b^{n} = b^{n}

, final term. Formula dono endpoints cover karta hai, sirf interior nahin.

Agar

T_{r + 1}

mein

r > n

ya negative

r

try karein toh

Finite expansion ke liye yeh meaningless hai. Hum define karte hain

(r n) = 0

jab bhi integer

r > n

r < 0

ho, kyunki literally zero tarike hain jitne brackets exist karte hain se zyada choose karne ke (ya negative number choose karne ke). Toh aisa koi term arise nahin hota; counter strictly

r = 0, 1, \dots, n

hai.

(a + b)^{n}

ka same-sign check

(1 - 1)^{n}

ke liye kaam karta hai

Haan —

a = 1, b = - 1

set karne par

\sum (r n) (- 1)^{r} = 0

milta hai

n \geq 1

ke liye, jo dikhata hai ki alternating coefficients exactly cancel ho jaate hain. Ek neat sanity test.

Jab

n

odd ho toh kitne "middle" terms hote hain aur kyun

Do — kyunki

n + 1

even hota hai, toh koi single term centre par nahin baith sakta; pair

T_{(n + 1) /2}

aur

T_{(n + 3) /2}

middle ko straddle karte hain.

Kya exponent equation

n - (something) r = k

non-integer

r

de sakti hai, aur iska kya matlab hai

Haan, aur iska matlab hai aisa koi term exist nahin karta —

r

{0, \dots, n}

mein ek whole number hona chahiye, toh fractional solution bolta hai ki

x

ki requested power kabhi appear nahin hoti.

Recall Jaane se pehle ek-line self-test

Teen jagah batao jahan counter $r$ tumhe trip kar sakta hai. Answer ::: (1) term position $r + 1$ hai, $r$ nahin; (2) $r$ $[0, n]$ mein integer hona chahiye; (3) $a$ ka exponent $n - r$ hai, jo $r$ badhne par ghatata hai.

Connections

Binomial theorem — expansion, general term — woh parent jise yeh bank stress-test karta hai.
Combinations nCr — har "why $(r n)$ " trap ke peeche counting engine.
Permutations vs Combinations — order-doesn't-matter confusions resolve karta hai.
Factorials — $(r n) = \frac{n !}{r ! ( n - r )!}$ computation traps.
Pascal's Triangle — symmetry aur Pascal-rule "why" questions.
Binomial Probability Distribution — same coefficient shape ek random experiment mein.