Ex 8.2,10 - Chapter 8 Class 11 Binomial Theorem

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Ex8.2, 10 The coefficients of the (r 1)th, rth and (r + 1)th terms in the expansion of (x + 1)n are in the ratio 1 : 3 : 5. Find n and r. Finding (r 1)th term , rth & (r + 1)th term of (x + 1)n Writing (x + 1)n as (1 + x)n We know that general term of expansion (a + b)n is Tr+1 = nCr an rbr For (1 + x)n , Putting a = 1 , b = a Tr+1 = nCr 1n r xr Tr+1 = nCr xr Coefficient of (r + 1)th term = nCr For rth term of (1 + x)n Replacing r with r 1 in (1) Tr 1 + 1 = nCr 1 xr 1 Tr = nCr 1 xr 1 Coefficient of (r)th term = nCr 1 For (r 1)th term of (1 + x)n Replacing r with r 2 in (1) Tr 2 + 1 = nCr 2 xr 2 Tr 1 = nCr 2 xr 2 Coefficient of (r 1)th term = nCr-2 Since the coefficient of (r 1)th, rth and (r + 1)th terms are in ratio 1 : 3 : 5 ( 1) = 1 3 2 1 = 1 3 ! 2 ![ 2) ! ! 1 ! ( 1) ! = 1 3 ! 2 ![ 2) ! 1 ![ 1) ! ! = 1 3 ! 1 2 ![ 1) ! ! 2 ![ 2) ! = 1 3 1 1 ! ( 2 )! = 1 3 1 + 1 ! ( + 2)! = 1 3 1 + 1 ! ( + 2)( +2 1)! = 1 3 1 + 1 ! ( + 2)( +1)! = 1 3 1 ( + 2) = 1 3 3(r 1) = n r + 2 3r 3 = n + 2 r n + 2 r 3r + 3 = 0 n 4r + 5 = 0 Also ( + 1) = 3 5 ! 1 ![ 1) ! ! ! ! = 3 5 ! 1 !( +1)! ! ! ! = 3 5 ! 1 ! ! ! 1 ! ( +1)! = 3 5 ! ( +1)! = 3 5 ( )! ( +1) ( )! = 3 5 + 1 = 3 5 5r = 3 (n r + 1) 5r = 3n 3r + 3 0 = 3n 3r + 3 5r 0 = 3n 8r + 3 3n 8r + 3 = 0 Now our equations are n 4r + 5 = 0 (1) & 3n 8r + 3 = 0 (2) From (1) n 4r + 5 = 0 n = 4r 5 Putting n = 4r 5 in (2) 3n 8r + 3 = 0 3(4r 5) 8r + 3 = 0 12r 15 8r + 3 = 0 12r 8r 15 + 3 = 0 4r 12 = 0 4r = 12 r = 12 4 r = 3 Putting r = 3 in n = 4r 5 n = 4(3) 5 n = 12 5 n = 7 Hence n = 7 & r = 3