find the coefficient of {x^5} in the expansion of{(1+2x)^6}{(1-x)^7}

 ncert cbse chapter 8 binomial theorem miscellaneous exercise 

 3.find the coefficient of {x^5} in the expansion of{(1+2x)^6}{(1-x)^7}

using binomial expansion

 (1+2x)^6 = 1 +C(6,1)(2x)+C(6,2)(2x)^2+C(6,3)(2x)^3 +C(6,4)(2x)^4 

+C(6,5)(2x)^5+C(6,6)(2x)^6

 C(6,1) = 6 =C(6,5)

C(6,2)=15=C(6,4)

C(6,3)=20

C(6,6)=1

(1+2x)^6 =

 1 +6(2x) +15(4(x^2))+20(8(x^3))+15(16(x^4))+6(32(x^5))+1(32(x^6))

=1+12x+60(x^2)+160(x^3)+240(x^4)+192(x^5)+32(x^6)

 

(1-x)^7=

1-C(7,1)(x)+C(7,2)(x^2)-C(7,3)(x^3)+C(7,4)(x^4)

-C(7,5)(x^5)+C(7,6)(x^6)-C(7,7)(x^7) 

C(7,1)=7=C(7,6) 

C(7,2)=21=C(7,5) 

C(7,3)=35=C(7,4)  

C(7,7)=1

 (1-x)^7=1 -7x+21(x^2)-35(x^3)+35(x^4)-21(x^5)+7(x^6)-1(x^7)

 

now multiplying the two expansions and concentrate only on the terms containing

(x^5)

1* {-21(x^5)}+{+12x}*{+35(x^4}+{+60(x^2)}*{-35(x^3)}+{+160(x^3)}{+21(x^2)}+{+240(x^4)}{-7x}+{+192(x^5)}{1}

 

coefficient of  (x^5) is

1*(-21)+(12)*(35)+(60)(-35)+(160)(21)+(240)(-7) +192*1

=(-21)+420-2100+3360-1680+192

=171

 

chapter 8 binomial theorem miscellaneous exercise

 1.Find a , b and n in the expansion of (a+b)^n if the first three terms in the expansion are 729, 7290, 30375

solution

 

2.  Find a if the coefficients of (x^2)  & (x^3) in the expansion of {(3+ax)^9} are equal 

solution

  3.find the coefficient of {x^5} in the expansion of{(1+2x)^6}{(1-x)^7}

solution

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