Binomial theorem — expansion, general term: Exercises

2.7.12 · D4 · Maths › Statistics & Probability — Intermediate › Binomial theorem — expansion, general term

Shuru karne se pehle, counting number $(r n)$ ke baare mein ek reminder — yeh har answer ka engine hai.

Level 1 — Recognition

L1.1 $(3 x - 5)^{11}$ ke expansion mein kitne terms hain?

Recall Solution

Ek binomial $(a + b)^{n}$ hamesha == $n + 1$ == terms mein expand hota hai — ek har value of $r$ ke liye, $0$ se $n$ tak. Yahan $n = 11$ hai, toh count hai $n + 1 = 11 + 1 = 12 .$ Humne kya kiya: humne kuch bhi expand nahi kiya — bas $n$ padha aur $1$ add kar diya. Bracket ke andar ki cheezein ( $3 x$ , $- 5$ ) terms ki sankhya ko kabhi affect nahi karti.

L1.2 $(a + b)^{7}$ mein, jis term mein $r = 3$ ho, usमें $a$ aur $b$ ke exponents likho.

Recall Solution

General term hai $T_{r + 1} = (r 7) a^{7 - r} b^{r}$ . $r = 3$ ke saath:

$b$ ka exponent hai $r = 3$ ,
$a$ ka exponent hai $n - r = 7 - 3 = 4$ .

Check karo ki dono ka sum $n$ ho: $4 + 3 = 7 = n$ . ✓ Term hai $(3 7) a^{4} b^{3} = 35 a^{4} b^{3}$ .

Level 2 — Application

L2.1 $(1 + 2 x)^{6}$ mein $x^{3}$ ka coefficient dhundo.

Recall Solution

General term: $T_{r + 1} = (r 6) (1)^{6 - r} (2 x)^{r} = (r 6) 2^{r} x^{r}$ . Hum chahte hain ki $x$ ki power $3$ ho, toh $r = 3$ : $coefficient = (3 6) 2^{3} = 20 \times 8 = 160 .$ $2^{r}$ kyun include karna hai? Kyunki $b = 2 x$ mein number $2$ hai; numeric part $2^{r}$ counting number $(r 6)$ ko multiply karta hai. Coefficient kabhi bhi sirf $(r n)$ nahi hota jab letter ke saath koi number ho.

L2.2 $(2 + x)^{4}$ ka poora expansion likho.

Recall Solution

$T_{r + 1} = (r 4) 2^{4 - r} x^{r}$ use karo aur $r$ ko $0 \to 4$ tak chalao:

$r$	$(r 4)$	$2^{4 - r}$	term
0	1	16	$16$
1	4	8	$32 x$
2	6	4	$24 x^{2}$
3	4	2	$8 x^{3}$
4	1	1	$x^{4}$

$(2 + x)^{4} = 16 + 32 x + 24 x^{2} + 8 x^{3} + x^{4} .$ Humne kya kiya: yahan $a = 2$ (ek number) aur $b = x$ hai. Toh numeric factor $2^{4 - r}$ ghatta hai jaise $r$ badhta hai, kyunki $a$ ki power decrease hoti hai.

L2.3 $(x + 3)^{8}$ ke expansion mein $4$ tha term dhundo.

Recall Solution

$4$ tha term matlab $T_{r + 1}$ jahan $r + 1 = 4$ , yani $r = 3$ (counter term number se ek peeche hota hai): $T_{4} = (3 8) x^{8 - 3} 3^{3} = 56 \cdot x^{5} \cdot 27 = 1512 x^{5} .$ $r = 3$ kyun? Kyunki $T_{1}$ case $r = 0$ hai. Toh $4$ tha term hamesha $r = 3$ use karta hai.

Level 3 — Analysis

L3.1 $(x^{3} + \frac{2}{x})^{6}$ mein $x^{10}$ wala term dhundo.

Recall Solution

General term: $T_{r + 1} = (r 6) (x^{3})^{6 - r} (\frac{2}{x})^{r} = (r 6) 2^{r} x^{3 (6 - r)} x^{- r} = (r 6) 2^{r} x^{18 - 4 r} .$ Exponents kyun combine karte hain? Taaki poora term $x$ ki ek power ban jaaye — sirf wohi cheez jise hum ek target ke barabar set kar sakte hain. Exponent ko $10$ set karo: $18 - 4 r = 10 \Rightarrow 4 r = 8 \Rightarrow r = 2.$ $T_{3} = (2 6) 2^{2} x^{10} = 15 \cdot 4 x^{10} = 60 x^{10} .$

L3.2 $(2 x^{2} - \frac{1}{x})^{9}$ mein $x$ se independent term (constant term) dhundo.

Recall Solution

General term: $T_{r + 1} = (r 9) (2 x^{2})^{9 - r} (\frac{- 1}{x})^{r} = (r 9) 2^{9 - r} (- 1)^{r} x^{2 (9 - r) - r} .$ $x$ ka exponent hai $2 (9 - r) - r = 18 - 3 r$ . " $x$ se independent" ka matlab yeh exponent $0$ ho: $18 - 3 r = 0 \Rightarrow r = 6.$ $T_{7} = (6 9) 2^{3} (- 1)^{6} = 84 \cdot 8 \cdot 1 = 672 .$ $(- 1)^{r}$ kyun rakhte hain? Kyunki $b = - \frac{1}{x}$ negative hai. Kyunki $r = 6$ even hai, $(- 1)^{6} = + 1$ , toh yahan sign positive rehta hai — lekin agar $r$ odd hota toh flip ho jaata. Sign kabhi mat chodo.

Level 4 — Synthesis

L4.1 $(3 x - \frac{1}{2 x})^{8}$ ka middle term dhundo.

Recall Solution

Kyunki $n = 8$ even hai, sirf ek middle term hai position $\frac{n}{2} + 1 = 5$ par, jo $r = 4$ use karta hai. $T_{5} = (4 8) (3 x)^{8 - 4} (\frac{- 1}{2 x})^{4} = (4 8) 3^{4} x^{4} \cdot \frac{( - 1 ) ^{4}}{2 ^{4} x ^{4}} .$ Numerator aur denominator mein $x^{4}$ cancel ho jaata hai, ek constant bachta hai: $T_{5} = 70 \cdot 81 \cdot \frac{1}{16} = \frac{5670}{16} = \frac{2835}{8} .$ Humne kya combine kiya: middle-term rule ( $n$ even ⇒ ek middle term $\frac{n}{2} + 1$ par) aur ek aise term ki algebra jisme $x$ -powers cancel ho jaati hain. $\frac{2835}{8} = 354.375$ .

L4.2 $(1 + x)^{n}$ mein, $x^{2}$ aur $x^{3}$ ke coefficients equal hain. $n$ dhundo.

Recall Solution

$(1 + x)^{n}$ mein $x^{r}$ ka coefficient exactly $(r n)$ hota hai (koi numeric factors nahi, kyunki dono letters "clean" hain: $1$ aur $x$ ). Dono coefficients equal set karo: $(2 n) = (3 n) .$ Factorials se likho aur simplify karo (dono sides ko divide karo — sabse aasaan hai ratio): $\frac{( 3 n )}{( 2 n )} = \frac{n - 2}{3} = 1 \Rightarrow n - 2 = 3 \Rightarrow n = 5 .$ Check: $(2 5) = 10$ aur $(3 5) = 10$ . ✓ Equal, jaise required. Humne kya synthesise kiya: L2 se identity " $(1 + x)^{n}$ mein $x^{r}$ ka coefficient $(r n)$ hai" aur Combinations nCr se factorial ratio $(r + 1 n) / (r n) = \frac{n - r}{r + 1}$ .

Level 5 — Mastery

L5.1 Product $(1 + x)^{4} (1 - x)^{6}$ mein $x^{5}$ ka coefficient dhundo.

Recall Solution

Strategy: hamen har term ki zaroorat nahi — sirf woh tarike chahiye jisme $x$ ki do chosen powers multiply hokar $x^{5}$ den. $(1 + x)^{4}$ mein $x^{i}$ ka coefficient $(i 4)$ likho, aur $(1 - x)^{6}$ mein $x^{j}$ ka $(j 6) (- 1)^{j}$ . Hamen chahiye $i + j = 5$ , jahan $0 \leq i \leq 4$ aur $0 \leq j \leq 6$ .

$i$	$j = 5 - i$	$(i 4)$	$(j 6) (- 1)^{j}$	product
0	5	1	$6 \cdot (- 1) = - 6$	$- 6$
1	4	4	$15 \cdot (+ 1) = 15$	$60$
2	3	6	$20 \cdot (- 1) = - 20$	$- 120$
3	2	4	$15 \cdot (+ 1) = 15$	$60$
4	1	1	$6 \cdot (- 1) = - 6$	$- 6$

Products add karo: $- 6 + 60 - 120 + 60 - 6 = - 12 .$ Humne kya kiya: ek product mein $x^{5}$ ka coefficient saare splits $i + j = 5$ ka (coeff of $x^{i}$ ) $\times$ (coeff of $x^{j}$ ) ka sum hai. Har negative sign $(1 - x)^{6}$ ke andar ke $(- 1)^{j}$ se aaya — ise rakhna zaruri tha.

L5.2 $(1 + k x)^{n}$ ke pehle teen terms ( $k$ constant, $n$ positive integer) hain $1 + 12 x + 60 x^{2} + \dots .$ $k$ aur $n$ dhundo.

Recall Solution

$(1 + k x)^{n}$ ko general term $T_{r + 1} = (r n) (k x)^{r} = (r n) k^{r} x^{r}$ se expand karo:

$r = 0$ : constant $= 1$ ✓ (automatically),
$r = 1$ : $x$ ka coefficient hai $(1 n) k = nk$ ,
$r = 2$ : $x^{2}$ ka coefficient hai $(2 n) k^{2} = \frac{n ( n - 1 )}{2} k^{2}$ .

Diye gaye expansion se match karo: $nk = 12 aur \frac{n ( n - 1 )}{2} k^{2} = 60.$ $k$ eliminate karo. Pehli equation se $k = \frac{12}{n}$ . Doosri mein substitute karo: $\frac{n ( n - 1 )}{2} \cdot \frac{144}{n ^{2}} = 60 \Rightarrow \frac{72 ( n - 1 )}{n} = 60.$ $72 (n - 1) = 60 n \Rightarrow 72 n - 72 = 60 n \Rightarrow 12 n = 72 \Rightarrow n = 6.$ Phir $k = \frac{12}{6} = 2$ . $n = 6, k = 2.$ $x^{2}$ term check karo: $\frac{6 \cdot 5}{2} \cdot 2^{2} = 15 \cdot 4 = 60$ . ✓

L5.3 $(1 + x)^{10}$ ke expansion mein greatest coefficient dhundo, aur batao woh kis term mein hai.

Recall Solution

$(1 + x)^{10}$ mein coefficients hain $(0 10), (1 10), \dots, (10 10)$ — Pascal's Triangle ki $10$ vi row. Yeh middle tak ek single peak tak badhte hain aur symmetrically girte hain. Even $n = 10$ ke liye peak $r = \frac{n}{2} = 5$ par hai: $(5 10) = \frac{10 !}{5 ! 5 !} = 252.$ Yeh $x^{5}$ ka coefficient hai, yaani $T_{6}$ term (kyunki $r = 5 \Rightarrow r + 1 = 6$ ). $Greatest coefficient = 252, in T_{6} = 252 x^{5} .$ Middle kyun? Kyunki $\frac{( r + 1 10 )}{( r 10 )} = \frac{10 - r}{r + 1}$ $r < 4.5$ ke liye $> 1$ hai (coefficients abhi badh rahe hain) aur $r > 4.5$ ke liye $< 1$ (girte hain). Turnover exactly $r = 5$ par hota hai. Neeche figure dekho.

Figure — Binomial theorem — expansion, general term

Active recall

Recall Quick self-test (pehle predict karo, phir reveal karo)

$(2 x^{2} - \frac{1}{x})^{9}$ ka constant term kaunsa $r$ deta hai? ::: $18 - 3 r = 0 \Rightarrow r = 6$ . $(3 x - 5)^{11}$ mein kitne terms hain? ::: $12$ . $(1 + 2 x)^{6}$ mein $x^{3}$ ka coefficient? ::: $(3 6) 2^{3} = 160$ . $(1 + x)^{10}$ ke middle term ki position? ::: $6$ tha term, $r = 5$ , coefficient $252$ . $(1 + x)^{n}$ mein, agar $x^{2}$ aur $x^{3}$ ke coeffs equal hain, toh $n =$ ? ::: $5$ .

Connections

Binomial theorem — expansion, general term — parent: jahan se $T_{r + 1}$ aata hai.
Combinations nCr — ratio $(r + 1 n) / (r n) = \frac{n - r}{r + 1}$ jo L4/L5 mein use hota hai.
Pascal's Triangle — row- $n$ coefficients aur unka single peak.
Factorials — har $(r n)$ kaise compute hota hai.
Binomial Probability Distribution — same coefficients ek random experiment ko power karte hain.

Exercises — Binomial theorem — expansion, general term

Level 1 — Recognition

Level 2 — Application

Level 3 — Analysis

Level 4 — Synthesis

Level 5 — Mastery

Active recall

Connections

Solution Map