BINOMIAL COEFFICIENTS BINOMIAL COEFFICIENTS BINOMIAL COEFFICIENTS Theorem If

BINOMIAL COEFFICIENTS

BINOMIAL COEFFICIENTS

BINOMIAL COEFFICIENTS Theorem : If cr denotes ncr then i) C 0 +C 1 +C 2 + …. . +Cn = 2 n ii) C 0 +C 2 +C 4+…… = C 1 +C 3 +C 5 + …. = 2 n-1 Proof We know that (1 + x)n = C 0 + C 1 x + C 2 x 2 + …. + Cn xn (1) Take x = 1 in (1), We get 2 n= C 0 + C 1 + C 2 + …. + Cn (2) Take x = – 1 in (1), we get 0 = C 0 – C 1 + C 2 – C 3…+ (-1)n Cn (3)

BINOMIAL COEFFICIENTS (2) + (3) C 0 + C 1 + C 2 + …. + Cn = 2 n C 0 – C 1 + C 2 – C 3…+ (-1)n Cn = 0 C 0 + C 1 + C 2 + …. + Cn = 2 n (2) n C 2(-1) + Cn 4 C+ =…. . ]=2 C 0 – C 1 + C 22[C – C 03+ …+ 0 n (3) =2 n-1 (4) (2) - (3) C 0 + C 1 + C 2 + …. + Cn = 2 n C 0 – C 1 + C 2 – C 3…+ (-1)n Cn = 0 2[C 1 + C 3 + C 5 + …. . ]=2 n

BINOMIAL COEFFICIENTS =2 n-1 (5) C 0 + C 2 + C 4 +. . . = C 1 + C 3 + C 5 +. . . = 2 n-1

BINOMIAL COEFFICIENTS NOTE

BINOMIAL COEFFICIENTS Theorem: If Cr denotes n. Cr , then We know +that na. C 0 + (a + d). C 1 +(a + 2 d). C 2 + …. +(a +nd). n. Cn = (2 a nd). 2 n Cr = Cn-r 1 Proof Let S = a. C 0 + (a + d). C 1 + (a + 2 d). C 2 + …. +(a + nd). Cn (1) On writing R. H. S of equation (1) in the reverse order, we get S = (a + nd). Cn + [a + (n – 1)d]. Cn-1 + …. (a + d). C 1 +a. C 0 S=(a + nd). C 0 + [a +(n – 1)d]. C 1+…. (a + d). Cn-1 +a. Cn (2) [∵ C 0 = Cn , C 1= Cn-1 …. . ]

BINOMIAL COEFFICIENTS (1)+(2) Take ‘(2 a + nd) ’ as common S = a. C 0 + (a + d). C 1 + (a + 2 d). C 2 + …. [a +(n – 1)d]. Cn-1 +(a + nd). Cn is nothing S =(a + nd). C 0 + [a +(n –This 1)d]. C 1 +…. but (a + d). Cn-1 +a. Cn ‘ 2 n’ Þ 2 S = (2 a + nd). C 0 +(2 a +nd). C 1 +…… +(2 a + nd). Cn-1 +(2 a + nd). Cn = (2 a + nd) (C 0 + C 1 +C 2 + …… + Cn) 2 S = (2 a + nd) 2 n S = a. C 0 + (a + d). C 1 + (a + 2 d). C 2 + …. +(a + nd). Cn (1) S=(a + nd). C 0 + [a +(n – 1)d]. C 1 +…. (a + d). Cn-1 +a. Cn (2)

BINOMIAL COEFFICIENTS S = (2 a + nd). 2 n-1

BINOMIAL COEFFICIENTS Theorem: If Cr denotes n. Cr , then C 1 + 2. C 2 x + 3. C 3 x 2 + …. +n. Cnxn-1 = n(1 + x)n-1 Proof We know that C 0 + C 1 x + C 2 x 2 + …. + Cn xn = (1 + x)n (1) Diff. (1) with respect to ‘x’ on both sides ⇒ C 1. 1 + C 2. 2 x + C 3. 3 x 2 + …. + Cn. nxn-1 = n(1 + x)n - 1

BINOMIAL COEFFICIENTS If x = 1 in (2), then we get i) C 1 + 2. C 2 + 3. C 3 + …+n. Cn = n. 2 n-1 If x = -1 in (2), then we get ii) C 1 – 2. C 2 + 3. C 3 – ……. . = 0 Note

BINOMIAL COEFFICIENTS Since C 1 + 2. C 2 x + 3. C 3 x 2 + …. +n. Cnxn-1 = n(1 + x)n-1 Diff. with respect to ‘x’ on both sides, we get 2. 1. C 2 + 3. 2. C 3 x + 4. 3. C 4 x 2 + …. + n. (n – 1). Cn xn-2= n(n – 1)(1 + x)n-2 Note

BINOMIAL COEFFICIENTS Multiply and divide with (n+1)x Proof L. H. S =

BINOMIAL COEFFICIENTS = (n+1)C 0+ (n+1)C 1 x+ (n+1)C 2 x 2+ (n+1)C (n+1)x This is nothing but (1+x)n+1 = R. H. S 3 x 3+…+ (n+1)C (n+1)x n+1 - (n+1)C 0

BINOMIAL COEFFICIENTS Note Theorem : If Cr denotes n. Cr , then C 0 Cr + C 1 Cr+1 + C 2 Cr+2 + …. + Cn-r Cn = 2 n. Cn+r. Proof We know that (1 + x)n = C 0 + C 1 x + C 2 x 2 +…+ Crxr+ Cr+1 xr+1 +…. + Cnxn (1)

BINOMIAL COEFFICIENTS (x + 1)n = C 0 xn + C 1 xn-1 + C 2 xn-2 + …. +Cr-1 xn-r+1 + Cr xn-r +…. + Cn (2) Multiplying (1) and (2) we get ; (C 0 + C 1 x + C 2 x 2 +…+ Crxr+ Cr+1 xr+1 +…. + Cnxn ) (C 0 xn + C 1 xn-1 + C 2 xn-2+…. +Cr-1 xn-r+1 + Cr xn-r +…. + Cn ) = (1 +x)n (x +1)n = (1 + x)2 n Comparing the coefficient of xn+r on both sides, we get

BINOMIAL COEFFICIENTS C 0 Cr + C 1 Cr+1 + C 2 Cr+2 + … + Cn-r Cn = 2 n. Cn+r (3) If r = 0, then from (3), we get C 02+ C 12 + C 22 +. . . + Cn 2 = 2 n. Cn If r = 1, then from (3), we get C 0 C 1 + C 1 C 2 + C 2 C 3 + …. . + Cn-1 Cn = 2 n. Cn+1

BINOMIAL COEFFICIENTS EXERCISE 6. 3 PROBLEMS

BINOMIAL COEFFICIENTS VERY SHORT ANSWER PROBLEMS 1) Prove that C 0 + 2. C 1 + 22. C 2 + ……+2 n. Cn = 3 n. Solution We know that (1 + x)n = C 0 + C 1 x + C 2 x 2 + …. + Cn xn If x = 2, then (1 + 2)n = C 0 + C 1. 2 + C 2. 22 + C 3. 23 + …. + Cn. 2 n

BINOMIAL COEFFICIENTS LONG ANSWER PROBLEMS Write the equation 1) Show that C 0 + 3. C 1 + 5. C 2 + …. + (2 n + 1). (2 n + 2). 2 n-1. n =reverse (1)Cin order Solution Let S = 1. C 0 + 3. C 1 + 5. C 2 + … + (2 n– 1). Cn– 1 + (2 n+1). Cn (1) S = (2 n+1). Cn + (2 n– 1). Cn– 1 + …… + 3. C 1 + 1. C 0 = (2 n+1). C 0 + (2 n– 1). C 1 +…. + 3. Cn– 1 + 1. Cn (2) [∵ C 0 = Cn , C 1= Cn-1 …. . Cr=cn-r]

BINOMIAL COEFFICIENTS (1) + (2) S = 1. C 0 + 3. CTake C 2 + …as + (2 n– 1). Cn– 1 + (2 n+1). Cn 1 + 5. (2 n+2) S = (2 n+1). C 0 +common (2 n– 1). C 1 +…. + 3. Cn– 1 + 1. Cn 1. C 0 + 3. C 1 + 5. C 2 + … + (2 n– 1). Cn– 1 + (2 n+1). Cn ⇒ 2 S = (2 n+2). C 0+ (2 n+2). C 1 +…. + (2 n+2). Cn– 1 + (2 n+2). Cn (1) (2 n+1). C + (2 n– 1). C +…. + 3. C + 1. C (2) 0 n– 1 ⇒ 2 S = (2 n + 2) [C 0 + C 1 1+ … + Cn– 1 + C n] n = (2 n+2) 2 n– 1

BINOMIAL COEFFICIENTS 2) Show that C 0 – 4. C 1 + 7. C 2 – 10. C 3 + … = 0. Solution L. H. S = 1. C 0 – 4. C 1 + 7. C 2 – 10. C 3 + …

BINOMIAL COEFFICIENTS = 3 (0) + 0 =0 = R. H. S.

BINOMIAL COEFFICIENTS Solution

BINOMIAL COEFFICIENTS = n(n-1). 2(n-2) +n. 2(n-1) = n(n-1). 2(n-2) +n. 2. 2(n-2) = n. 2(n-2){n-1+2} = n(n+1). 2(n-2) = R. H. S

BINOMIAL COEFFICIENTS Solution L. H. S = = n+(n-1)+(n-2)+. . . +1 =∑n =R. H. S

BINOMIAL COEFFICIENTS 5) Show that 3. C 02+7. C 12+11. C 22+. . (4 n+3). Cn 2=(2 n+3)2 n. Cn ∵ n. Cr=n. Cn-r Solution n C =n C 0 n n C =n C 1 n-1 On writing R. H. S of equation (1) in the reverse order, we get

BINOMIAL COEFFICIENTS (1) + (2) 2 S= 3. C 02 Take +7. C(4 n+6) Cn-12 +(4 n+3). Cn 2 as 1 +. . +(4 n-1). common S=(4 n+3). C 02+(4 n-1). C 12 +. . . +7. Cn-12+3. Cn 2 2 S=(4 n+6). C 02+(4 n+6). C 12 +. . . +(4 n+6)Cn-12+(4 n+6)Cn 2 2 S = (4 n+6)[C 02+C 12+. . . +Cn-12+Cn 2] (2 n. Cn) = (2 n+3) 2 n. Cn 3. C 02+7. C 12+11. C 22+. . (4 n+3). Cn 2=(2 n+3)2 n. Cn

BINOMIAL COEFFICIENTS 7) If (1 + x 2)n = a 0 + a 1 x + a 2 x 2 +……+a 2 nx 2 n, then show that i) a 0 + a 1 + a 2 + …… +a 2 n = 3 n Solution Given (1 + x 2)n = a 0 + a 1 x + a 2 x 2 +……+a 2 nx 2 n (1) i) Put x = 1 in eq. (1) Þ (1 + 1)n = a 0 + a 1(1) + a 2 (1)2 + …. + a 2 n(1)2 n

BINOMIAL COEFFICIENTS Put x = – 1 in (1) (1– 1+1)n = a 0 + a 1 (– 1) a 2 (– 1)2 + …. + a 2 n(– 1)2 n (1 + x 2)n = a 0 + a 1 x + a 2 x 2 +……+a 2 nx 2 n (1) n (2) + (3) a 0 + a 1 + a 2 + a 3 + …. + a 2 n = 3 a 0 – a 1 + a 2 – a 3 + …. + a 2 n = 1 2 a 0 + 2 a 2 + 2 a 4 + …. + 2 a 2 n = 3 n + 1 2[a 0 + a 2 + a 4 + …. + a 2 n]= 3 n + 1

BINOMIAL COEFFICIENTS Solution (2) – (3) 2 a 1 + 2 a 3 + 2 a 5 + …. + 2 a 2 n-1 = 3 n – 1 2[a 1 + a 3 + a 5 + …. + a 2 n– 1] = 3 n – 1

BINOMIAL COEFFICIENTS iv) a 0 + a 3 + a 6 + a 9 + … = 3 n 1 Solution Put x = in eq. (1) 1 + + 2 = 0 3 = 1 a 0 + a 1 ( ) + a 2 2 + a 3 3 + …. +a 2 n 2 n = (1 + + 2)n a 0 + a 1 + a 2 2 + a 3. 1 + …. +a 2 n 2 n = 0 (4) (1 + x 2)n = a 0 + a 1 x + a 2 x 2 +……+a 2 nx 2 n (1) Put x = 2 in eq. (1) a 0 + a 1 2 + a 2 4 +a 3 6 + …. + a 2 n( 2)2 n = (1 + 2 + 4)n a 0 + a 1 22 n+a 2 + a 3. 1 + …. + a 2 n 4 n =2 n 0 (5) 2 (1 + x ) = a 0 + a 1 x + a 2 x +……+a 2 nx (1)

BINOMIAL COEFFICIENTS (2) + (4) + (5) a 0 + a 1 + a 2 + a 3 + …. + a 2 n = 3 n a 0 + a 1 + a 2 2 + a 3. 1 + …. +a 2 n 2 n = 0 a 0 + a 1 2 +a 2 + a 3. 1 + …. + a 2 n 4 n = 0 Þ 3 a 0 + a 1 (1 + + 2) + a 2 (1 + + 2) + 3 a 3 + ……+a 2 n (1 + 2 n + 4 n) = 3 n+ 0 3. a 0 + 0 + 3 a 3 + …. = 3 n 3[a 0 + a 3 + a 6 + …. . ] = 3 n

BINOMIAL COEFFICIENTS Solution L. H. S = Multiply and divide with ‘n+1’

BINOMIAL COEFFICIENTS = R. H. S

BINOMIAL COEFFICIENTS Solution L. H. S = (C 0+C 1) (C 1+C 2). . (Cn-1+Cn)

BINOMIAL COEFFICIENTS = R. H. S

BINOMIAL COEFFICIENTS Solution (1 + x)n = C 0 + C 1 x + C 2 x 2 +…. . +Cn xn (1) (x – 1)n = C 0 xn – C 1 xn-1 + ……. + (– 1)n Cn (2) Multiply (1) & (2) (1 + x)n (x – 1)n =(C 0 + C 1 x + C 2 x 2 +…. . +Cn xn). (C 0 xn – C 1 xn-1 + ……. + (– 1)n Cn)

BINOMIAL COEFFICIENTS (x 2 – 1)n =(C 0 + C 1 x + C 2 x 2 +…. . +Cn xn). (C 0 xn – C 1 xn-1 + – ……. + (– 1)n Cn) (3) In L. H. S of (3), Tr+1 = n. Cr (x 2)n-r. (-1)r = (-1)r n. Cr x 2 n-2 r For the coefficient of xn, Take n = 2 n – 2 r

BINOMIAL COEFFICIENTS Hence coefficient of xn in (x 2 – 1)n is n. Cn/2(-1)n/2 If ‘n’ is odd then If n is even then

BINOMIAL COEFFICIENTS Thank you…