Intuition The big pictureWhen you expand ( a + b ) n (a+b)^n ( a + b ) n n n n ( a + b ) (a+b) ( a + b ) choosing either a a a b b b b b b k k k a a a n − k n-k n − k a n − k b k a^{n-k}b^k a n − k b k How many ways can you make that choice? Exactly ( n k ) \binom{n}{k} ( k n ) a n − k b k a^{n-k}b^k a n − k b k ( n k ) \binom{n}{k} ( k n ) whole theorem in one sentence.
Definition Binomial Theorem (positive integer index)For any real/complex a , b a,b a , b n n n ( a + b ) n = ∑ k = 0 n ( n k ) a n − k b k (a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{\,n-k} b^{\,k} ( a + b ) n = ∑ k = 0 n ( k n ) a n − k b k binomial coefficient is
( n k ) = n ! k ! ( n − k ) ! , 0 ≤ k ≤ n . \binom{n}{k} = \frac{n!}{k!\,(n-k)!}, \qquad 0 \le k \le n. ( k n ) = k ! ( n − k )! n ! , 0 ≤ k ≤ n .
Number of terms in the expansion = = = n + 1 = = = ==n+1== === n + 1 ==
The general (i.e. ( k + 1 ) (k+1) ( k + 1 ) T k + 1 = ( n k ) a n − k b k T_{k+1} = \binom{n}{k} a^{n-k} b^k T k + 1 = ( k n ) a n − k b k
Powers of a a a decrease from n n n 0 0 0 b b b increase from 0 0 0 n n n n n n
Intuition Feynman-style "why the coefficient is( n k ) \binom{n}{k} ( k n ) ( a + b ) n = ( a + b ) ( a + b ) ⋯ ( a + b ) ⏟ n brackets (a+b)^n = \underbrace{(a+b)(a+b)\cdots(a+b)}_{n\text{ brackets}} ( a + b ) n = n brackets ( a + b ) ( a + b ) ⋯ ( a + b ) n n n b k b^k b k b b b k k k Choosing which k k k is ( n k ) \binom{n}{k} ( k n ) a n − k b k a^{n-k}b^k a n − k b k
This is the fastest intuition (80/20), but exams want the induction proof , so we derive it rigorously below.
We need one algebraic engine first.
Claim P ( n ) P(n) P ( n ) ( a + b ) n = ∑ k = 0 n ( n k ) a n − k b k \displaystyle (a+b)^n = \sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^k ( a + b ) n = k = 0 ∑ n ( k n ) a n − k b k n ≥ 1 n\ge 1 n ≥ 1
Base case n = 1 n=1 n = 1 RHS = ( 1 0 ) a 1 b 0 + ( 1 1 ) a 0 b 1 = a + b = ( a + b ) 1 = \binom{1}{0}a^1b^0 + \binom{1}{1}a^0b^1 = a+b = (a+b)^1 = ( 0 1 ) a 1 b 0 + ( 1 1 ) a 0 b 1 = a + b = ( a + b ) 1
Inductive step. Assume P ( n ) P(n) P ( n ) ( a + b ) (a+b) ( a + b ) ( a + b ) n + 1 = ( a + b ) ∑ k = 0 n ( n k ) a n − k b k . (a+b)^{n+1} = (a+b)\sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^k. ( a + b ) n + 1 = ( a + b ) ∑ k = 0 n ( k n ) a n − k b k .
Distribute:
= ∑ k = 0 n ( n k ) a n − k + 1 b k + ∑ k = 0 n ( n k ) a n − k b k + 1 . = \sum_{k=0}^{n}\binom{n}{k}a^{n-k+1}b^k \;+\; \sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^{k+1}. = ∑ k = 0 n ( k n ) a n − k + 1 b k + ∑ k = 0 n ( k n ) a n − k b k + 1 .
Why this step? We split into an "a a a b b b
Re-index the second sum with j = k + 1 j=k+1 j = k + 1 k = j − 1 k=j-1 k = j − 1 j j j 1 → n + 1 1\to n+1 1 → n + 1 ∑ j = 1 n + 1 ( n j − 1 ) a n − j + 1 b j . \sum_{j=1}^{n+1}\binom{n}{j-1}a^{n-j+1}b^{j}. ∑ j = 1 n + 1 ( j − 1 n ) a n − j + 1 b j . j → k j\to k j → k a ( n + 1 ) − k b k a^{(n+1)-k}b^{k} a ( n + 1 ) − k b k
Combine. Pull out the k = 0 k=0 k = 0 ( n 0 ) a n + 1 = a n + 1 \binom{n}{0}a^{n+1}=a^{n+1} ( 0 n ) a n + 1 = a n + 1 k = n + 1 k=n+1 k = n + 1 ( n n ) b n + 1 = b n + 1 \binom{n}{n}b^{n+1}=b^{n+1} ( n n ) b n + 1 = b n + 1 1 ≤ k ≤ n 1\le k\le n 1 ≤ k ≤ n ( a + b ) n + 1 = a n + 1 + ∑ k = 1 n [ ( n k ) + ( n k − 1 ) ] a ( n + 1 ) − k b k + b n + 1 . (a+b)^{n+1} = a^{n+1} + \sum_{k=1}^{n}\Big[\binom{n}{k}+\binom{n}{k-1}\Big]a^{(n+1)-k}b^{k} + b^{n+1}. ( a + b ) n + 1 = a n + 1 + ∑ k = 1 n [ ( k n ) + ( k − 1 n ) ] a ( n + 1 ) − k b k + b n + 1 .
Apply Pascal's rule ( n k ) + ( n k − 1 ) = ( n + 1 k ) \binom{n}{k}+\binom{n}{k-1}=\binom{n+1}{k} ( k n ) + ( k − 1 n ) = ( k n + 1 ) a n + 1 = ( n + 1 0 ) a n + 1 a^{n+1}=\binom{n+1}{0}a^{n+1} a n + 1 = ( 0 n + 1 ) a n + 1 b n + 1 = ( n + 1 n + 1 ) b n + 1 b^{n+1}=\binom{n+1}{n+1}b^{n+1} b n + 1 = ( n + 1 n + 1 ) b n + 1 ( a + b ) n + 1 = ∑ k = 0 n + 1 ( n + 1 k ) a ( n + 1 ) − k b k . (a+b)^{n+1} = \sum_{k=0}^{n+1}\binom{n+1}{k}a^{(n+1)-k}b^{k}. ( a + b ) n + 1 = ∑ k = 0 n + 1 ( k n + 1 ) a ( n + 1 ) − k b k .
That is exactly P ( n + 1 ) P(n+1) P ( n + 1 ) P ( n ) P(n) P ( n ) n ≥ 1 n\ge 1 n ≥ 1 ■ \blacksquare ■
Worked example 2 — A specific term (no full expansion)Find the coefficient of x 6 x^{6} x 6 ( 2 x 2 − 1 x ) 9 \left(2x^2 - \dfrac{1}{x}\right)^{9} ( 2 x 2 − x 1 ) 9 T k + 1 = ( 9 k ) ( 2 x 2 ) 9 − k ( − 1 x ) k = ( 9 k ) 2 9 − k ( − 1 ) k x 2 ( 9 − k ) − k T_{k+1}=\binom{9}{k}(2x^2)^{9-k}\left(-\tfrac1x\right)^{k} = \binom9k 2^{9-k}(-1)^k x^{2(9-k)-k} 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 Why this step? We collect the power of x x x x 18 − 3 k x^{18-3k} x 18 − 3 k 18 − 3 k = 6 ⇒ k = 4 18-3k = 6 \Rightarrow k=4 18 − 3 k = 6 ⇒ k = 4 = ( 9 4 ) 2 5 ( − 1 ) 4 = 126 ⋅ 32 = 4032 =\binom94 2^{5}(-1)^4 = 126\cdot 32 = \boxed{4032} = ( 4 9 ) 2 5 ( − 1 ) 4 = 126 ⋅ 32 = 4032
b = 1 b=1 b = 1 ( 1 + x ) n (1+x)^n ( 1 + x ) n a = 1 , b = x a=1,\ b=x a = 1 , b = x ( 1 + x ) n = ∑ k = 0 n ( n k ) x k (1+x)^n=\sum_{k=0}^n\binom nk x^k ( 1 + x ) n = ∑ k = 0 n ( k n ) x k x = 1 x=1 x = 1 ∑ k = 0 n ( n k ) = 2 n \sum_{k=0}^n\binom nk = 2^n ∑ k = 0 n ( k n ) = 2 n n n n x = − 1 x=-1 x = − 1 ∑ k = 0 n ( − 1 ) k ( n k ) = 0 \sum_{k=0}^n(-1)^k\binom nk = 0 ∑ k = 0 n ( − 1 ) k ( k n ) = 0 n ≥ 1 n\ge1 n ≥ 1 Why: these are just the theorem evaluated at handy points — pure consequences, no new work.
Common mistake Forgetting powers must sum ton n n Wrong feels right: writing ( a + b ) 3 = a 3 + b 3 (a+b)^3 = a^3+b^3 ( a + b ) 3 = a 3 + b 3 feels symmetric.
Fix: every term is a n − k b k a^{n-k}b^k a n − k b k n n n ( a + b ) 3 = a 3 + 3 a 2 b + 3 a b 2 + b 3 (a+b)^3=a^3+3a^2b+3ab^2+b^3 ( a + b ) 3 = a 3 + 3 a 2 b + 3 a b 2 + b 3 n + 1 = 4 n+1=4 n + 1 = 4 2 2 2
Common mistake Dropping the sign inside the binomialIn ( a − b ) n (a-b)^n ( a − b ) n ( − b ) (-b) ( − b ) T k + 1 = ( n k ) a n − k ( − 1 ) k b k T_{k+1}=\binom nk a^{n-k}(-1)^k b^k T k + 1 = ( k n ) a n − k ( − 1 ) k b k ( − 1 ) k (-1)^k ( − 1 ) k Fix: always substitute the whole second term (with its sign) into b b b
Common mistake Confusing "k k k ( k + 1 ) (k+1) ( k + 1 ) T k + 1 = ( n k ) a n − k b k T_{k+1}=\binom nk a^{n-k}b^k T k + 1 = ( k n ) a n − k b k k k k k k k 0 0 0 Fix: to get the r r r k = r − 1 k=r-1 k = r − 1
Common mistake Bad re-indexing in the proofWhen shifting j = k + 1 j=k+1 j = k + 1 limits . If k k k 0 → n 0\to n 0 → n j j j 1 → n + 1 1\to n+1 1 → n + 1
Recall Feynman: explain to a 12-year-oldImagine n n n a a a b b b a a b aab aab b a b bab bab k k k b b b k k k b b b n n n k k k ( a + b ) n (a+b)^n ( a + b ) n
Mnemonic Remember the term"Coefficient chooses, first-letter falls, second-letter rises."
( n k ) \binom nk ( k n ) ⋅ a n − k \cdot\; a^{n-k} ⋅ a n − k ⋅ b k \cdot\; b^{k} ⋅ b k = n + 1 = n+1 = n + 1 "one more than the power."
#flashcards/maths
State the binomial theorem for positive integer n n n ( a + b ) n = ∑ k = 0 n ( n k ) a n − k b k (a+b)^n=\sum_{k=0}^n\binom nk a^{n-k}b^k ( a + b ) n = ∑ k = 0 n ( k n ) a n − k b k ( n k ) = n ! k ! ( n − k ) ! \binom nk=\frac{n!}{k!(n-k)!} ( k n ) = k ! ( n − k )! n !
How many terms are in the expansion of ( a + b ) n (a+b)^n ( a + b ) n n + 1 n+1 n + 1
Write the general term T k + 1 T_{k+1} T k + 1 T k + 1 = ( n k ) a n − k b k T_{k+1}=\binom nk a^{n-k}b^k T k + 1 = ( k n ) a n − k b k
State Pascal's rule. ::: ( n k − 1 ) + ( n k ) = ( n + 1 k ) \binom{n}{k-1}+\binom nk=\binom{n+1}{k} ( k − 1 n ) + ( k n ) = ( k n + 1 )
Which lemma powers the inductive step of the proof? ::: Pascal's rule, to merge the two shifted sums.
In the induction, what substitution aligns the second sum? ::: j = k + 1 j=k+1 j = k + 1 a n + 1 − k b k a^{n+1-k}b^k a n + 1 − k b k
What is ∑ k = 0 n ( n k ) \sum_{k=0}^n\binom nk ∑ k = 0 n ( k n ) 2 n 2^n 2 n a = b = 1 a=b=1 a = b = 1
What is ∑ k = 0 n ( − 1 ) k ( n k ) \sum_{k=0}^n(-1)^k\binom nk ∑ k = 0 n ( − 1 ) k ( k n ) n ≥ 1 n\ge1 n ≥ 1 0 0 0 a = 1 , b = − 1 a=1,b=-1 a = 1 , b = − 1
Coefficient of x 6 x^6 x 6 ( 2 x 2 − 1 / x ) 9 (2x^2-1/x)^9 ( 2 x 2 − 1/ x ) 9 4032 4032 4032 k = 4 k=4 k = 4
Why is the coefficient of a n − k b k a^{n-k}b^k a n − k b k ( n k ) \binom nk ( k n ) b b b k k k n n n
multiply by (a+b), split, re-index
Binomial Theorem statement
Intuition Hinglish mein samjhoDekho, binomial theorem ka core idea bilkul simple hai: ( a + b ) n (a+b)^n ( a + b ) n ( a + b ) (a+b) ( a + b ) n n n a a a b b b b b b k k k n n n k k k b b b ( n k ) \binom{n}{k} ( k n ) ( n k ) a n − k b k \binom{n}{k}a^{n-k}b^k ( k n ) a n − k b k
Proof induction se hota hai. Base case n = 1 n=1 n = 1 n n n ( a + b ) (a+b) ( a + b ) a a a b b b j = k + 1 j=k+1 j = k + 1 Pascal's rule ( n k − 1 ) + ( n k ) = ( n + 1 k ) \binom{n}{k-1}+\binom{n}{k}=\binom{n+1}{k} ( k − 1 n ) + ( k n ) = ( k n + 1 ) n + 1 n+1 n + 1
Exam me general term T k + 1 = ( n k ) a n − k b k T_{k+1}=\binom{n}{k}a^{n-k}b^k T k + 1 = ( k n ) a n − k b k x x x 18 − 3 k 18-3k 18 − 3 k k k k ( a − b ) n (a-b)^n ( a − b ) n b b b ( − b ) (-b) ( − b ) ( − 1 ) k (-1)^k ( − 1 ) k
Bottom line: binomial theorem = smart counting + Pascal's rule se induction. Ye aage probability, series sums, aur approximations sab me kaam aata hai, isliye foundation strong rakho.
