Binomial theorem — expansion, general term: Worked examples

2.7.12 · D3 · Maths › Statistics & Probability — Intermediate › Binomial theorem — expansion, general term

The scenario matrix

Har binomial-expansion question actually kuch independent "knobs" ka combination hota hai. Yahan saare knobs ka full set hai aur woh kya values le sakte hain:

Knob (kya vary karta hai)	Possible values	Kaun sa example hit karta hai
Sign of $b$	$b > 0$ (plain) / $b < 0$ (alternating signs)	Ex 1 ( $+$ ), Ex 3 ( $-$ )
Numeric factor inside $a, b$	none (baby case $1 + x$ ) / present ( $2 x$ , $3/ x^{2}$ )	Ex 1 (none), Ex 3 (both)
Goal type	full expansion / specific power $x^{k}$ / constant term / coefficient only	Ex 1, Ex 2, Ex 3, Ex 6
Parity of $n$	$n$ even (one middle term) / $n$ odd (two middle terms)	Ex 4 (even), Ex 5 (odd)
Degenerate exponent	no valid $r$ exists (term absent!)	Ex 7
Word / real-world	probability-flavoured story	Ex 8
Exam twist	two conditions at once (equal terms / ratio)	Ex 9
Non-standard $n$	$n = 0$ (trivial) / $n$ negative or fractional (infinite series)	Ex 10

Is table ko padhne ka rule: har row se ek value chuno aur tumne ek possible question describe kar liya. Neeche ke das examples is liye choose kiye gaye hain taaki har row ki har value kam se kam ek baar aaye.

Example 1 — baby case, $b > 0$ , full expansion · (Sign $+$ , no numeric factor, full expansion)

Is example se pehle, yahan wo coefficient tool hai jis par hum lean karte hain — Pascal's Triangle scratch se banaya hua:

Figure — Binomial theorem — expansion, general term

Figure (alt text): Pascal's Triangle ki pehli chhe rows ek pyramid ki tarah midnight-navy par draw ki gayi hain. Har number uske do diagonally upar wale numbers ka sum hota hai (do green arrows har interior entry mein neeche point karte hain). Row $0$ upar ek single $1$ hai; row $5$ neeche $15101051$ padhti hai — yellow mein highlight ki gayi hai kyunki Example 1 exactly usi row ka use karta hai. Edges sab $1$ 's hain (blue), kyunki $(0 n) = (n n) = 1$ .

Worked example

$(1 + x)^{5}$ ko completely expand karo. Forecast: aage padhne se pehle — kitne terms honge, aur middle term ka coefficient kya hoga? Apna guess likho.

Step 1. $a = 1$ , $b = x$ , $n = 5$ identify karo, toh $T_{r + 1} = (r 5) 1^{5 - r} x^{r} = (r 5) x^{r}$ . Yeh step kyun? Jab $a = 1$ , toh factor $1^{5 - r} = 1$ vanish ho jaata hai — yeh baby case hai jahan coefficient purely $(r 5)$ hota hai, kuch nahi. Yeh ek aisa case hai jahan "coefficient $= (r n)$ " poori tarah sach hai.

Step 2. Coefficients ko upar wale triangle ki row $5$ se padho (yellow highlight): $1, 5, 10, 10, 5, 1$ . Yeh step kyun? Pascal's Triangle ki row $n$ hi list $(0 n), \dots, (n n)$ hoti hai — jab row choti ho toh har factorial compute karne ki zaroorat nahi.

Step 3. $x$ ki badhti powers attach karo: $(1 + x)^{5} = 1 + 5 x + 10 x^{2} + 10 x^{3} + 5 x^{4} + x^{5} .$

Verify: $n + 1 = 6$ terms hain ✓. $x = 1$ set karo: left side $2^{5} = 32$ hai; right side $1 + 5 + 10 + 10 + 5 + 1 = 32$ hai ✓. ( $x = 1$ set karna saare coefficients sum karta hai — ek classic sanity check.)

Example 2 — find the term containing a target power · (power $x^{k}$ , numeric-free)

Example 3 — negative $b$ AND numeric factors AND constant term · (Sign $-$ , factors present, constant term)

Yeh ek saath teen matrix cells hit karta hai — "boss fight".

Worked example

$(2 x^{2} - \frac{1}{x})^{9}$ mein $x$ se independent term nikalo. Forecast: kya tumhara answer positive hoga ya negative? Ek guess likho — sign wahi jagah hai jahan zyaadatar log slip karte hain.

Step 1. $a = 2 x^{2}$ , $b = - \frac{1}{x}$ , $n = 9$ likho: $T_{r + 1} = (r 9) (2 x^{2})^{9 - r} (- \frac{1}{x})^{r} .$ Yeh step kyun? Poora signed $b$ , minus samete, substitute karo. $b$ ko $+ \frac{1}{x}$ treat karna $(- 1)^{r}$ silently drop kar dega.

Step 2. Har factor ko (number)(power of $x$ ) mein split karo: $T_{r + 1} = (r 9) 2^{9 - r} (- 1)^{r} x^{2 (9 - r)} x^{- r} = (r 9) 2^{9 - r} (- 1)^{r} x^{18 - 3 r} .$ Yeh step kyun? Saare $x$ 's ko ek exponent $18 - 3 r$ mein collect karte hain taaki isse target ke against set kar sakein. Numeric part $2^{9 - r} (- 1)^{r}$ saath chal raha hai.

Step 3. " $x$ se independent" ka matlab exponent $0$ hai: $18 - 3 r = 0 \Rightarrow r = 6$ . Yeh step kyun? Constant term $x^{0}$ hota hai; solve karne se woh unique bracket-count milta hai jo $x$ mita deta hai.

Step 4. $T_{7} = (6 9) 2^{3} (- 1)^{6} = 84 \cdot 8 \cdot 1 = 672.$ Yeh step kyun? $(6 9) = (3 9) = 84$ ; $2^{9 - 6} = 2^{3} = 8$ ; $(- 1)^{6} = + 1$ (even power ⇒ positive).

Verify: exponent check $18 - 3 (6) = 0$ ✓. Sign: $r = 6$ even hai, toh $(- 1)^{6} = + 1$ , answer positive $= 672$ ✓.

Example 4 — middle term, $n$ EVEN · (parity even)

Worked example

$(x - \frac{2}{x})^{6}$ ka middle term nikalo. Forecast: even $n = 6$ — ek middle term hai ya do? Kaun sa index?

Step 1. $a = x$ , $b = - \frac{2}{x}$ , $n = 6$ identify karo, toh general term hai $T_{r + 1} = (r 6) x^{6 - r} (- \frac{2}{x})^{r} .$ Yeh step kyun? Hamesha pehle $a$ , $b$ , $n$ name karo aur general term likho, exactly Ex 1–3 ki tarah. Signed $b = - \frac{2}{x}$ abhi substitute karna guarantee karta hai ki minus automatically carry ho jaata hai.

Step 2. $n = 6$ even hai, toh position $\frac{n}{2} + 1 = 4$ par ek single middle term hai, yaani $r = 3$ . Yeh step kyun? $n + 1 = 7$ terms ke saath, exact centre 4th hai; $T_{r + 1}$ mein $r + 1 = 4$ se $r = 3$ milta hai.

Step 3. $T_{4} = (3 6) x^{6 - 3} (- \frac{2}{x})^{3} = (3 6) x^{3} \cdot (- 2)^{3} x^{- 3} .$ Yeh step kyun? Signed $- \frac{2}{x}$ ko intact rakho; $(- 2)^{3}$ negative hoga.

Step 4. $T_{4} = 20 \cdot (- 8) \cdot x^{0} = - 160.$ Yeh step kyun? $(3 6) = 20$ , $(- 2)^{3} = - 8$ , aur $x^{3} x^{- 3} = x^{0} = 1$ .

Verify: $(3 6) = \frac{6 \cdot 5 \cdot 4}{6} = 20$ ✓. Sign: $r = 3$ odd ⇒ $(- 2)^{3} < 0$ ⇒ negative $= - 160$ ✓.

Example 5 — middle term, $n$ ODD (DO hain) · (parity odd)

Worked example

$(2 + x)^{7}$ ke do middle terms nikalo. Forecast: odd $n = 7$ se $8$ terms milte hain — beech mein kaun se do baithe hain?

Step 1. $n + 1 = 8$ terms (even count) ke saath, koi single centre nahi hai; do middle terms 4th aur 5th hain, yaani $r = 3$ aur $r = 4$ . Yeh step kyun? Jab $n$ odd hota hai toh terms ki sankhya even hoti hai, isliye centre do terms ke beech padta hai — dono "middle" hain. Positions $\frac{n + 1}{2} = 4$ aur $\frac{n + 3}{2} = 5$ hain.

Step 2. $T_{4} = (3 7) 2^{7 - 3} x^{3} = 35 \cdot 16 x^{3} = 560 x^{3}$ . Yeh step kyun? Yahan $a = 2$ numeric weight $2^{n - r}$ carry karta hai; isse bhool jao aur coefficient galat ho jaayega.

Step 3. $T_{5} = (4 7) 2^{7 - 4} x^{4} = 35 \cdot 8 x^{4} = 280 x^{4}$ . Yeh step kyun? $(4 7) = (3 7) = 35$ ; power $2^{3} = 8$ shrink hoti hai jaise $r$ badhta hai kyunki $a$ ka exponent shrink hota hai.

Verify: $(3 7) = 35$ ✓. $2^{4} = 16 \Rightarrow 35 \cdot 16 = 560$ ✓; $2^{3} = 8 \Rightarrow 35 \cdot 8 = 280$ ✓.

Example 6 — sirf coefficient, factor present · (coefficient-only goal, numeric factor)

Example 7 — degenerate case: term EXIST NAHI KARTA · (no valid $r$ )

Matrix mein ek nasty cell hai jo zyaadatar notes skip karte hain: agar $r$ ki equation ka koi whole-number solution nahi hai? Toh requested term simply expansion mein nahi hoga.

Worked example Kya

$(x^{3} + \frac{1}{x})^{8}$ mein $x^{5}$ ka term hai? Forecast: haan ya nahi? Solve karne se pehle answer dene ki koshish karo.

Step 1. $T_{r + 1} = (r 8) (x^{3})^{8 - r} x^{- r} = (r 8) x^{24 - 3 r - r} = (r 8) x^{24 - 4 r}$ . Yeh step kyun? Ek single power $24 - 4 r$ mein collapse karo taaki target exponent dhundh sakein.

Step 2. $24 - 4 r = 5 \Rightarrow 4 r = 19 \Rightarrow r = \frac{19}{4} = 4.75$ set karo. Yeh step kyun? $r$ zaroor ${0, 1, \dots, 8}$ mein se ek whole number hona chahiye kyunki yeh brackets count karta hai. Fraction impossible hai.

Step 3. Kyunki $r = 4.75$ integer nahi hai, koi $x^{5}$ term nahi hai — iska coefficient $0$ hai. Yeh step kyun? Is expansion mein har exponent $24 - 4 r$ form ka hai, yaani yeh $4$ ki steps mein girta hai: $24, 20, 16, 12, 8, 4, 0, \dots$ — $5$ kabhi nahi aata.

Verify: $r = 0..8$ ke liye reachable exponents ${24, 20, 16, 12, 8, 4, 0, - 4, - 8}$ hain; $5 \in /$ is set mein ✓, toh answer "exist nahi karta" sahi hai.

Example 8 — real-world / probability flavour · (word problem)

Binomial coefficients sirf algebra nahi hain — yeh count karte hain. Yeh example wahi $(r n)$ engine use karta hai jo Binomial Probability Distribution ko power karta hai.

Worked example Ek fair coin

$6$ baar toss ki jaati hai. $(H + T)^{6}$ ki expansion ko ek bookkeeping device ki tarah use karte hue, $2^{6}$ equally-likely outcomes mein se kitne mein exactly $2$ heads hain? Forecast: $64$ total outcomes mein se count guess karo.

Step 1. Symbolically expand karo: $(H + T)^{6} = \sum_{r = 0}^{6} (r 6) H^{6 - r} T^{r}$ . Yeh step kyun? Har toss ek bracket se $H$ ya $T$ choose karta hai — exactly "switch" picture. Term $H^{6 - r} T^{r}$ un saari sequences ko collect karta hai jinmein $6 - r$ heads aur $r$ tails hain.

Step 2. "Exactly $2$ heads" ka matlab $6 - r = 2$ , toh $r = 4$ (chaar tails). Matching term $H^{2} T^{4}$ hai jiska coefficient $(4 6) = 15$ hai. Yeh step kyun? Hum count sahi term ke coefficient se padhna chahte hain, $H^{2} T^{4}$ , jiska index $r = 4$ hai. Woh coefficient $(4 6)$ count karta hai ki $6$ tosses mein se kaun se $4$ tails hain.

Step 3. Equivalently, seedha count karo ki kaun se $2$ tosses heads hain: $(2 6) = 15$ . Yeh step kyun? Symmetry $(4 6) = (2 6)$ se — $4$ tails choose karna wahi hai jaise bacha hua $2$ heads choose karna. Dono counts same $15$ dete hain, toh "heads count karo" aur "tails count karo" agree karte hain. Isliye students bina galti ke $(2 6)$ likh sakte hain.

Step 4. (Bonus) probability $= \frac{15}{2 ^{6}} = \frac{15}{64}$ . Yeh step kyun? $64$ sequences mein se har ek equally likely hai, aur $15$ favourable hain.

Verify: total sequences $\sum_{r} (r 6) = 2^{6} = 64$ ✓ ( $H = T = 1$ set karo). Favourable $(4 6) = (2 6) = 15$ ✓.

Example 9 — exam twist: ek saath do conditions · (equal-coefficients twist)

Worked example

$(1 + x)^{n}$ mein $x^{r}$ aur $x^{r + 1}$ ke coefficients equal hain, aur yeh $n = 15$ par $r = 7$ ke liye hota hai. Isse verify karo, aur general rule nikalo. Forecast: $n$ aur $r$ ke beech kaun sa relation do neighbouring coefficients ko match karne par majboor karta hai?

Step 1. $x^{r}$ ka coefficient $(r n)$ hai; $x^{r + 1}$ ka $(r + 1 n)$ hai. Inhe equal set karo. Yeh step kyun? Yeh baby case hai, toh coefficients pure binomials hain — twist dono mein se do ko compare karna hai.

Step 2. Ratio $\frac{( r + 1 n )}{( r n )} = \frac{n - r}{r + 1}$ use karo. Coefficients ko equal set karne ka matlab hai yeh ratio $= 1$ hai: $\frac{n - r}{r + 1} = 1 \Rightarrow n - r = r + 1 \Rightarrow n = 2 r + 1.$ Yeh step kyun? Consecutive binomial coefficients ka ratio sabse cleanest tool hai — yeh ek factorial equation ko ek one-line linear equation mein badal deta hai.

Step 3. Toh equality exactly tab hoti hai jab $n$ odd ho aur $r = \frac{n - 1}{2}$ : do middle coefficients equal hote hain. Yeh step kyun? $n = 2 r + 1$ precisely woh odd case hai jahan do middle terms exist karte hain (Ex 5) — woh symmetry $(r n) = (r + 1 n)$ se same coefficient share karte hain.

Step 4. Diye gaye numbers check karo: $n = 15 \Rightarrow r = \frac{15 - 1}{2} = 7$ ✓, aur $(7 15) = (8 15) = 6435$ .

Verify: $(7 15) = 6435$ aur $(8 15) = 6435$ equal hain ✓; $n = 2 (7) + 1 = 15$ ✓.

Example 10 — non-standard $n$ : trivial $n = 0$ aur infinite series · (non-standard index)

Parent note mein theorem ne assume kiya tha ki $n$ ek positive integer hai. Do edge cases us se bahar hain: trivial $n = 0$ , aur negative/fractional $n$ jahan expansion ek infinite series ban jaati hai — Taylor & Maclaurin Series ka gateway.

Worked example (a)

$(a + b)^{0}$ expand karo. (b) $(1 + x)^{- 1}$ aur $(1 + x)^{1/2}$ ke pehle teen terms likho, $∣ x ∣ < 1$ ke liye valid. Forecast: har ek mein kitne terms hain — finite ya endless?

Step 1 (trivial case). $(a + b)^{0} = 1$ kisi bhi $a, b$ ke liye (dono zero nahi). Yeh step kyun? $n = 0$ ke saath exactly $n + 1 = 1$ term hota hai: $(0 0) a^{0} b^{0} = 1$ . Sum turant ruk jaata hai — sabse chhoti possible expansion.

Step 2 (finite formula kyun fail hota hai). Ek positive integer $n$ ke liye, factor $(n - r)$ eventually $r = n + 1$ par $0$ ho jaata hai, jo baad ke har term ko kill kar deta hai — isi wajah se sum $n$ par ruk jaata hai. Jab $n$ negative ya fractional hota hai, $(n - r)$ kabhi $0$ nahi hota, toh koi term kabhi kill nahi hoti: series forever chalti rehti hai. Yeh step kyun? Yeh exact mechanism point karta hai: finiteness ek factor ke zero hone se aati hai, aur non-integer $n$ us zero ko hata deta hai.

Step 3 (generalised coefficient). Kisi bhi real $n$ ke liye coefficient falling product use karta hai (upar defined), factorials nahi: $(r n) = \frac{n ( n - 1 ) ( n - 2 ) \dots ( n - r + 1 )}{r !} .$ Yeh step kyun? Usual $(r n) = \frac{n !}{r ! ( n - r )!}$ ko $(n - r)!$ chahiye, jo fractional $n$ ke liye meaningless hai (koi " $(- \frac{3}{2})!$ " nahi hota). Top ko falling product se replace karne par $(n - r)!$ completely hata jaata hai, toh coefficient ab bhi sense karta hai — aur whole $n$ ke liye pehle jaisa number deta hai.

Step 4 ( $(1 + x)^{- 1}$ compute karo, $n = - 1$ ). Falling-product coefficients hain $(0 - 1) = 1$ , $(1 - 1) = \frac{- 1}{1} = - 1$ , $(2 - 1) = \frac{( - 1 ) ( - 2 )}{2} = 1$ . $x$ ki powers attach karo: $(1 + x)^{- 1} = 1 - x + x^{2} - \dots (∣ x ∣ < 1) .$ Yeh step kyun? $x$ ki powers ab bhi $x^{0}, x^{1}, x^{2}, \dots$ exactly finite case ki tarah chadhti hain; sirf coefficients ab falling product se aate hain.

Step 5 ( $(1 + x)^{1/2}$ compute karo, $n = \frac{1}{2}$ ). Coefficients $(0 1/2) = 1$ , $(1 1/2) = \frac{1/2}{1} = \frac{1}{2}$ , $(2 1/2) = \frac{\frac{1}{2} \cdot ( - \frac{1}{2} )}{2} = - \frac{1}{8}$ . Isliye $(1 + x)^{1/2} = 1 + \frac{1}{2} x - \frac{1}{8} x^{2} + \dots (∣ x ∣ < 1) .$ Yeh step kyun? Wahi recipe — top par falling product, neeche $r!$ — ab $n = \frac{1}{2}$ par apply ki gayi.

Step 6 ( $∣ x ∣ < 1$ kyun?). Series mein infinitely many terms hain, toh yeh function ke barabar tab hoti hai jab woh terms kuch nahi ban jaayein. Powers $x^{r}$ sirf tab shrink hoti hain jab $∣ x ∣ < 1$ ho (e.g. $0. 5^{r} \to 0$ ); agar $∣ x ∣ \geq 1$ terms khatam nahi hote aur endless sum ki koi finite value nahi hoti. Toh expansion sirf $∣ x ∣ < 1$ ke andar trustworthy hai — yeh infinitely many terms ki kimat hai, jo positive-integer case kabhi nahi chukta kyunki uska sum finite hota hai. Yeh step kyun? Ek finite sum hamesha valid hoti hai; ek infinite wali ko apne terms ka vanish hona zaroori hai. Woh region batana jahan aisa hota hai answer ka hissa hai, afterthought nahi.

Verify: $(1 + x)^{- 1} = \frac{1}{1 + x}$ jaani-pehchaani geometric series $1 - x + x^{2} - \dots$ hai ( $∣ x ∣ < 1$ ke liye valid) ✓. $1 + \frac{1}{2} x - \frac{1}{8} x^{2}$ ko square karne par $1 + x + O (x^{3})$ milta hai, jo $(1 + x)^{1/2}$ squared $= 1 + x$ se match karta hai ✓.

Poore matrix ki ek picture

Figure (alt text): ek graph jo $x$ ka exponent (vertical axis) ko counter $r$ (horizontal axis) ke against plot karta hai humare teen expansions ke liye. Blue line $18 - 3 r$ $(2 x^{2} \pm \frac{1}{x})^{9}$ hai (Ex 2/3); green line $24 - 4 r$ $(x^{3} + \frac{1}{x})^{8}$ hai (Ex 7). Ek dashed yellow horizontal line "constant term" target ( $0$ ) mark karti hai aur ek dashed red line target $x^{5}$ mark karti hai. Ek specific term padhna bas ek dot pick karna hai; " $x^{k}$ term dhundho" woh jagah dhundha hai jahan ek coloured line ek horizontal target cross kare; "aisa koi term nahi" (Ex 7) woh hai jab target height do dots ke beech baith jaaye aur kabhi kisi par na utare — red line kabhi kisi green dot ko touch nahi karti.

Active recall

Recall Forecast then verify

$(3 + 2 x)^{7}$ mein $x^{5}$ ka coefficient? ::: $6048$ .
Kya $(x^{3} + 1/ x)^{8}$ mein $x^{5}$ term hai? ::: Nahi — $r = 4.75$ integer nahi hai.
$(2 x^{2} - 1/ x)^{9}$ mein $x$ se independent term? ::: $672$ ( $r = 6$ par).
$6$ -toss outcomes ki sankhya jinmein exactly $2$ heads hain? ::: $(2 6) = (4 6) = 15$ .
$(1 + x)^{n}$ ke consecutive coefficients equal kab hote hain? ::: Jab $n = 2 r + 1$ (odd $n$ , do middle terms).
$(1 + x)^{- 1}$ ke pehle teen terms $∣ x ∣ < 1$ ke liye? ::: $1 - x + x^{2} - \dots$ .

Connections

Binomial theorem — expansion, general term — parent formula jo har example use karta hai.
Combinations nCr — kyun coefficient count karta hai (Ex 8).
Factorials — har $(r n)$ kaise compute hota hai (aur fractional $n$ ke liye kyun fail karte hain, Ex 10).
Pascal's Triangle — choti coefficient rows ka sabse fast route (Ex 1).
Binomial Probability Distribution — coin-toss example uska ek term hai.
Taylor & Maclaurin Series — non-integer $n$ ke liye Ex 10 ki infinite series.
Permutations vs Combinations — $(r n)$ ke peeche order-free choosing.