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Notes On Binomial Theorem 11th Class

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Published in: Mathematics
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BINOMIAL THEOREM Class 11th Maths : Binomial theorem for any positive integer n (a+b)^n

Abhinav K / Mumbai

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  1. CHAPTER 8 BINOMIAL THEOREM Binomial theorem for any positive integer n (a+b)n = n coan + n Clan 1b + n C2a b + C3an 3b3 + n-2 2 n ncnbn Recall 1) ncr= nC1 = 4) (n-r)! r! 2) ncr = ncn-r 8 6-8 x 7=28 = 35 Ix2x3 n OBSERVATIONS/ FORMULAS 1) The coefficients occurring in the binomial theorem are known as binomial coefficients. 2) There are (n+l) terms in the expansion of (a+b)n , ie one more than the index. 3) The coefficient of the terms equidistant from the beginning and end are equal. 4) (14-X)n = n CO + n CIX + n C2X2 + n C3X3 + b = x in the expansion of (a + b)n). 5) (I-xy = nco- + nc2X2 - nc3X3 + and b = -x in the expansion of (a + b)n). (4)) + ncnxn. (By putting a = 1 and + (-1 y ncnxn (By putting a = 1 (By putting x = 1 in +(- (By putting x = 1 in (5)) 8**) (r + term in the binomial expansion for (a+b)n is called the general term which is given by — an r bro i.e to find 4t term = T 4, substitute r
  2. 9* ) Middle term in the expansion of (a+b)n If n is even, middle term = If n is odd, then 2 middle terms are, th — + 1 term. th th + 1 term. term and 10*) To find the term independent of x or the constant term, find the coefficient of xo.(ie put power of x = 0 and find r) Problems Ex 8.1 (4 marks) Q (1 mark) (4 marks) 13* * 14* * (4 marks) 13**) Show that 9n+1 8n 9 is divisible by 64, whenever n is a positive integer Or 2n 4-2 - 8n - 9 is divisible by 64 Solution: 9n+1 8n - 9 = (1+8Y+l 8n — 9 CO + + n +1C28 + C 383 + 114-1 + n + ICn+18n+1 9 9 = 1 + 8 n + 8 + 8 [ C 2 + n+1C3.8 + (since = 8 114-1-2 + 8n l] which is divisible by 64 Problems = 8 [ C2 + n +1C3.8+ (4 marks) eg 8**,9** (6 marks)
  3. ? 1+X) JO 1101suedxo ? 1+1) pue “ SI (LOH ? '(IOH) ()' ? ?0H? 11'**01 ? ? ? ? I-I) ? ? 1+1-u) LP … € … I … … … … LIOA!D ? I) *8'(IOH)€'Z'OIJEUI 9 ? 1 0 ? u (I) 9 ? 01 0 1+ ? I-I) € + u 1-1091110S ? 1+1 -u XO J 1+1
  4. simplify as above and get the equation 3n 8r = -3 solving (1) and (2) we get n = 7 and r = 3. EXTRA/HOT QUESTIONS (2) 1) Using Binomial theorem show that 23n 7n 1 or 8n 7n 1 is divisible by 49 where n is a natural number. (4 marks**) 2) Find the coefficient of x3 in the equation of (1+2x)6 (I-x)7 (HOT) 3) Find n if the coefficient of 5th , & 7th terms in the expansion of (1+x)n are in A.P. 4) If the coefficient of x r 1, xr, xr+l in the expansion of (1+x)n are in A.P. prove that n2 (4r+1)n +4r2 O. (HOT) 5) If 6 , 7 ,8 & 9th terms in the expansion of (x+y)n are respectively a,b,c th th th &d then show that b2 ac = 4a (HOT) c2 — bd 3c 10 6) Find the term independent of x in the expansion of 3x (4 marks* ) 7) Using Binomial theorem show that 33n 26n — 1 is divisible by 676. marks ** ) 8) The 3rd,4th & terms in the expansion of (x+a)n are 84, 280 & 560 (4 respectively. Find the values of x, a and n. (6 marks**) 9) The coefficient of 3 consecutive terms in the expansion of (1+x)n the ratio 3 : 8 : 14. Find n. (6 mark**) are in 10) 11) 12) Find the constant term in the expansion of (x- l/x) Find the middle term(s) in the expansion of 14 a If (1+x) 10 x n = CO + CIX + C2X + Cnxn n = n.2n Prove that + 2C2 + 3C3 + Answers 2) -43 3) n = 7 or 14 6) 76545/8 8) x =1, a=2, n = 7 9) 10 10) -3432 11) i) -252 ii) -63 x14 32

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