General term of binomial expansion — finding specific terms

3.3.13 · Maths › Sequences & Series

WHY karte hain hum ek "general term" ka use?

WHAT chahiye humein: ek aisa formula jo $(a + b)^{n}$ ki koi bhi term deta ho ek index daalne par, taaki hum (maano) 7th term ya $x^{10}$ ka coefficient nikal sakein bina saare $n + 1$ terms likhe.

WHY ye matter karta hai: $(x + 2)^{15}$ ko poora expand karna matlab 16 terms ka dard. Agar question sirf "the coefficient of $x^{9}$ " poochhe, toh general term se hum ek line mein answer nikal sakte hain.

Isko scratch se derive karte hain

$(a + b)^{n} = n brackets (a + b) (a + b) \dots (a + b)$

HOW ek single term banti hai: product ki ek term banane ke liye, tum saare $n$ brackets se guzarte ho aur har ek mein se ya toh $a$ ya $b$ choose karte ho. Maano tumne exactly $r$ brackets se $b$ choose kiya. Tab:

woh $r$ brackets contribute karte hain $b^{r}$ ,
baaki $n - r$ brackets contribute karte hain $a^{n - r}$ .

Ye step kyun? Multiplication bas har bracket se ek term pick karta hai; product tumhari picks ka product hota hai.

Different ways ki sankhya, jisme tum choose karo ki $n$ brackets mein se kaun se $r$ brackets $b$ denge, woh hai $== (r n) ==$ . Un sab tarikhon se identical term $a^{n - r} b^{r}$ milti hai, isliye woh add hote hain:

$(all terms with r b’s) = (r n) a^{n - r} b^{r}$

Har possible $r$ ko $0$ se $n$ tak sum karne par:

$(a + b)^{n} = \sum_{r = 0}^{n} (r n) a^{n - r} b^{r}$

Figure — General term of binomial expansion — finding specific terms

Universal method (har baar kaam karta hai)

Kisi specific term ya coefficient dhundhne ke liye:

General term $T_{r + 1} = (r n) a^{n - r} b^{r}$ likhein apne actual $a, b, n$ ke saath.
Variable ki power ko collect karo ek exponent mein.
Woh exponent jo question maang raha hai uske barabar set karo → $r$ solve karo.
$r$ wapas plug karo term/coefficient paane ke liye.

Worked Example 1 — simple "kaun si term"

$(2 x + 3)^{7}$ ki 5th term dhundho.

5th term ka matlab $r + 1 = 5 \Rightarrow r = 4$ . Kyun? Term number hoti hai $r + 1$ .
Yahan $a = 2 x, b = 3, n = 7$ . $T_{5} = (4 7) (2 x)^{7 - 4} (3)^{4} = 35 \cdot (2 x)^{3} \cdot 81$ Ye step kyun? General term mein direct substitution. $= 35 \cdot 8 x^{3} \cdot 81 = 22680 x^{3}$

Worked Example 2 — specific power ka coefficient

$(x^{2} + \frac{1}{x})^{12}$ mein $x^{9}$ ka coefficient dhundho.

$a = x^{2}, b = x^{- 1}, n = 12$ . $T_{r + 1} = (r 12) (x^{2})^{12 - r} (x^{- 1})^{r} = (r 12) x^{2 (12 - r)} x^{- r}$ Kyun? Hum variable ki powers alag karte hain taaki unhe combine kar sakein. $= (r 12) x^{24 - 2 r - r} = (r 12) x^{24 - 3 r}$ Exponent ko 9 ke barabar set karo: $24 - 3 r = 9 \Rightarrow 3 r = 15 \Rightarrow r = 5$ 9 ke barabar kyun set kiya? Hum $x^{9}$ wali term chahte hain. $Coefficient = (5 12) = 792$

Worked Example 3 — $x$ se independent term (constant term)

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

$a = 2 x, b = - x^{- 2}, n = 9$ . $T_{r + 1} = (r 9) (2 x)^{9 - r} (- x^{- 2})^{r} = (r 9) 2^{9 - r} (- 1)^{r} x^{9 - r} x^{- 2 r}$ $= (r 9) 2^{9 - r} (- 1)^{r} x^{9 - 3 r}$ " $x$ se independent" ka matlab exponent $= 0$ : $9 - 3 r = 0 \Rightarrow r = 3$ 0 kyun? $x^{0} = 1$ mein koi $x$ nahi hota. $T_{4} = (3 9) 2^{6} (- 1)^{3} = 84 \cdot 64 \cdot (- 1) = - 5376$

Recall Feynman: ek 12-saal ke bacche ko samjhao

Socho tumhare paas $n$ boxes hain aur har box mein ya toh ek red ball ( $a$ ) ya blue ball ( $b$ ) chhupa hai. Ek "term" banane ke liye, tum har box kholte ho aur dekhte ho kya hai. Agar tum decide karo "mujhe exactly $r$ blue balls chahiye," toh tum $r$ blues aur $n - r$ reds ko multiply karte ho. Lekin kaun se boxes blue hain ye choose karne ke bahut tarike hain — woh count hai $(r n)$ . Toh har type ki term $(r n)$ baar aati hai: exactly yahi hai $(r n) a^{n - r} b^{r}$ . Koi khaas term grab karne ke liye, tum saare boxes nahi kholte — bas kehte ho "kitne blues se wo power milegi jo chahiye?" aur seedha wahan pahunch jaate ho.

Connections

Binomial Theorem for positive integer index
Binomial coefficients and Pascal's Triangle
Middle term of a binomial expansion
Greatest coefficient and greatest term
Combinations nCr
Factorials

(a + b)^{n}

ka general term

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

, jahan

r = 0, 1, \dots, n

T_{r + 1}

kyun hai,

T_{r}

kyun nahi?

Kyunki

r

start hota hai

0

se; pehli term

r = 0

se correspond karti hai, isliye

(r + 1)

-vi term index

r

use karti hai.

(a + b)^{n}

mein terms ki sankhya

n + 1

x^{k}

ka coefficient kaise dhundhein

T_{r + 1}

mein variable ka exponent collect karo, use

k

ke barabar set karo,

r

solve karo, phir coefficient evaluate karo.

x

se independent term ka matlab kya condition hai?

x

ki net power

0

ke barabar ho.

(x^{2} + 1/ x)^{12}

ke general term mein

x

ka exponent

24 - 3 r

(x^{2} + 1/ x)^{12}

mein

x^{9}

ka coefficient

(5 12) = 792

(

r = 5

se).

Middle term ka rule

n

even → ek middle term

T_{n /2 + 1}

;

n

odd → do middle terms

T_{(n + 1) /2}, T_{(n + 3) /2}

(2 x + 3)^{7}

ki 5th term

r = 4

T_{5} = (4 7) (2 x)^{3} (3)^{4} = 22680 x^{3}

Common sign trap in

(- 1/ x^{2})^{r}

(- 1)^{r} x^{- 2 r}

ki tarah likho; sign

r

ki parity par depend karti hai.