3.3.11
Jab tum ( a + b ) n n ( a + b ) choosing se aati hai — har bracket se ya toh a b k b n − k a a n − k b k Kitne tarike hain woh choice karne ke? Exactly ( k n ) a n − k b k ( k n ) poora theorem ek sentence mein.
Definition Binomial Theorem (positive integer index)Kisi bhi real/complex a , b n ( a + b ) n = ∑ k = 0 n ( k n ) a n − k b k binomial coefficient hai
( k n ) = k ! ( n − k )! n ! , 0 ≤ k ≤ n .
Expansion mein terms ki sankhya === n + 1 ==
General (yaani ( k + 1 ) T k + 1 = ( k n ) a n − k b k
a n 0 decrease hoti hain; b 0 n increase hoti hain; har term mein exponents ka sum n
Intuition Feynman-style "coefficient( k n ) ( a + b ) n = n brackets ( a + b ) ( a + b ) ⋯ ( a + b ) n b k k b Kaun se k — yeh ( k n ) a n − k b k
Yeh sabse tez intuition hai (80/20), lekin exams mein induction proof chahiye hoti hai, isliye hum neeche use rigorously derive karte hain.
Pehle humein ek algebraic engine chahiye.
Claim P ( n ) ( a + b ) n = k = 0 ∑ n ( k n ) a n − k b k n ≥ 1
Base case n = 1 RHS = ( 0 1 ) a 1 b 0 + ( 1 1 ) a 0 b 1 = a + b = ( a + b ) 1
Inductive step. Maano P ( n ) ( a + b ) ( a + b ) n + 1 = ( a + b ) ∑ k = 0 n ( k n ) a n − k b k .
Distribute karo:
= ∑ k = 0 n ( k n ) a n − k + 1 b k + ∑ k = 0 n ( k n ) a n − k b k + 1 .
Yeh step kyun? Hum ek "a b
Doosre sum ko re-index karo j = k + 1 k = j − 1 j 1 → n + 1 ∑ j = 1 n + 1 ( j − 1 n ) a n − j + 1 b j . j → k a ( n + 1 ) − k b k
Combine karo. Pehle sum se k = 0 ( 0 n ) a n + 1 = a n + 1 k = n + 1 ( n n ) b n + 1 = b n + 1 1 ≤ k ≤ n ( a + b ) n + 1 = a n + 1 + ∑ k = 1 n [ ( k n ) + ( k − 1 n ) ] a ( n + 1 ) − k b k + b n + 1 .
Pascal's rule apply karo ( k n ) + ( k − 1 n ) = ( k n + 1 ) a n + 1 = ( 0 n + 1 ) a n + 1 b n + 1 = ( n + 1 n + 1 ) b n + 1 ( a + b ) n + 1 = ∑ k = 0 n + 1 ( k n + 1 ) a ( n + 1 ) − k b k .
Yeh exactly P ( n + 1 ) P ( n ) n ≥ 1 ■
Worked example 2 — Ek specific term (full expansion nahi)( 2 x 2 − x 1 ) 9 x 6 T k + 1 = ( k 9 ) ( 2 x 2 ) 9 − k ( − x 1 ) k = ( k 9 ) 2 9 − k ( − 1 ) k x 2 ( 9 − k ) − k Yeh step kyun? Hum x x 18 − 3 k 18 − 3 k = 6 ⇒ k = 4 = ( 4 9 ) 2 5 ( − 1 ) 4 = 126 ⋅ 32 = 4032
b = 1 ( 1 + x ) n a = 1 , b = x ( 1 + x ) n = ∑ k = 0 n ( k n ) x k x = 1 ∑ k = 0 n ( k n ) = 2 n n x = − 1 ∑ k = 0 n ( − 1 ) k ( k n ) = 0 n ≥ 1 Kyun: yeh bas theorem ko convenient points pe evaluate karna hai — pure consequences hain, koi nayi mehnat nahi.
Common mistake Bhool jaana ki powers ka sumn Galat lagtaa sahi hai: ( a + b ) 3 = a 3 + b 3 symmetric lagta hai.
Fix: har term a n − k b k n ( a + b ) 3 = a 3 + 3 a 2 b + 3 a b 2 + b 3 n + 1 = 4 2
Common mistake Binomial ke andar sign chhod dena( a − b ) n ( − b ) T k + 1 = ( k n ) a n − k ( − 1 ) k b k ( − 1 ) k Fix: hamesha poora doosra term (apne sign ke saath) b
k ( k + 1 ) T k + 1 = ( k n ) a n − k b k k k 0 Fix: r k = r − 1
Common mistake Proof mein galat re-indexingj = k + 1 limits badlna bhool jaate hain. Agar k 0 → n j 1 → n + 1
Recall Feynman: ek 12-saal ke bachche ko explain karoSocho n a b aab bab k b k b n k ( a + b ) n
"Coefficient chooses, first-letter falls, second-letter rises."
( k n ) ⋅ a n − k ⋅ b k = n + 1 "power se ek zyada."
#flashcards/maths
Positive integer n ( a + b ) n = ∑ k = 0 n ( k n ) a n − k b k ( k n ) = k ! ( n − k )! n !
( a + b ) n n + 1 General term T k + 1 T k + 1 = ( k n ) a n − k b k
Pascal's rule state karo. ::: ( k − 1 n ) + ( k n ) = ( k n + 1 )
Proof ke inductive step ko kaun sa lemma power karta hai? ::: Pascal's rule, do shifted sums ko merge karne ke liye.
Induction mein doosre sum ko align karne ke liye kaun sa substitution use hota hai? ::: j = k + 1 a n + 1 − k b k
∑ k = 0 n ( k n ) 2 n a = b = 1 n ≥ 1 ∑ k = 0 n ( − 1 ) k ( k n ) 0 a = 1 , b = − 1 ( 2 x 2 − 1/ x ) 9 x 6 4032 k = 4 a n − k b k ( k n ) n k b
multiply by (a+b), split, re-index
Binomial Theorem statement
