SOLUTION: 1). Solve for r
7!/(7-r)! = 840
2). Expand (a-2)^5 using the binomal theoren.
(Is there a shorter way to expand this without multiplying (a-2) 5 times
and writin
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Question 23449: 1). Solve for r
7!/(7-r)! = 840
2). Expand (a-2)^5 using the binomal theoren.
(Is there a shorter way to expand this without multiplying (a-2) 5 times
and writing out a lot of numbers?)
a) a^5-8a^4+16a^3-24a^2+24a-32
b) a^5-10a^4+40a^3-80a^2+80a-32
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
1). Solve for r
7!/(7-r)! = 840
7!/840=(7-R)!
7*6*5*4*3*2*1/840=(7-R)!
3*2*1=(7-R)!=3!
7-R=3
R=7-3=4
2). Expand (a-2)^5 using the binomal theoren.
(Is there a shorter way to expand this without multiplying (a-2) 5 times
and writing out a lot of numbers?)
BINOMIAL THEOREM...
(X+A)^N=(NC0)*X^N+(NC1)*(X^(N-1))*(A)+(NC2)*(X^(N-2))*(A^2)+...+(NCR)X^(N-R)*(A^R)+.......+(NCN)*A^N
FOR NUMERICAL PROBLEMS YOU CAN USE THIS ORAL CALCULATION
(X+A)^5=1*X^5+(5*1/1)X^4*A+(5*4/2)X^3*A^2+(10*3/3)X^2*A^3+(10*2/4)X^1*A^4+(5*1/5)A^5
POWERS...START WITH X^5...OR...=X^5*A^0=...GO ON REDUCING BY 1 FOR EACH TERM
THE POWER OF X AND INCREASING BY 1 THE POWER OF A...THAT IS X^5,X^4*A,X^3*A^2....TILL YOU END WITH X^0*A^5=A^5
NOW COEFFICIENTS ARE CALCULATED AS SHOWN ABOVE AND EXPLAINED BELOW.
1.START WITH 1 FOR THE 1ST.TERM
2...2ND.TERM....MULTIPLY THE COEFFICIENT OF PREVIOUS TERM (1)WITH POWER OF X IN THAT (5)AND DIVIDE BY THE NUMBER OF TERMS OVER OR WRITTEN ALREADY(1)=5*1/1=5 IS THE COEFFICIENT OF NEXT OR NEW TERM
3.....3RD.TERM..SAME WAY REPEAT...MULTIPLY THE COEFFICIENT OF PREVIOUS TERM (5)WITH POWER OF X IN THAT (4)AND DIVIDE BY THE NUMBER OF TERMS OVER OR WRITTEN ALREADY(2)=5*4/2=10 IS THE COEFFICIENT OF NEXT OR NEW TERM
4..THE WORKING FOR OTHERS IS SHOWN ABOVE IN BRACKETS..
NOW YOU CAN DO YOUR PROBLEM USING
X=a AND A=-2
a) a^5-8a^4+16a^3-24a^2+24a-32
b) a^5-10a^4+40a^3-80a^2+80a-32
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