The sum of coefficients of the polynomial f(x) is equal to 2 and the sum of coefficients in even places is equal t the sum of cofficients in odd places.Find the remainder of dividing f(x) by qquad g(x)=x^(2)-1
Answers
Answer:
What is the sum of the coefficients of all the even powers of x in the expansion of (2x^-3x+1) ^11?
I guess the expression is (2x2–3x+1)11
If so:
(2x2–3x+1)11=((2x−1).(x−1))11=(2x−1)11.(x−1)11−−(A)
=[∑r=011(11r).2r.(−1)n−r.xr]∗[∑k=011(11k).(−1)n−k.xk]
=∑r=011∑k=011(11r).(11k).2r.(−1)n−r+n−k.xk+r
One can substitute the possible combinations for k and r to yield even powers of x, to arrive at the result.
Alternative method # 1:
To get even powers:
(a) Multiply coefficients of even powers of (2x−1)11∗ coefficients of even powers of (x−1)11
(b) Multiply coefficients of odd powers of (2x−1)11∗ coefficients of odd powers of (x−1)11
(c) sum of (a) and (b) would give the result
Alternative method # 2:
(2x2–3x+1)11=[2x2+(1−3x)]11=∑r=011(11r).(2x2)11−r.(1−3x)r
=∑r=011(11r).211−r.x22−2r.∑k=0r.(rk).1k.(−3x)r−k
=∑r=011∑k=0r(11r).(rk).211−r.(−3)r−k.x22−3r−k
For even powers of x, the possible values of r and k are, such that 22–3r−k is divisible by 2 and r≤k :
(0,0), (1,1),(2,0),(2,2), (3,1),(3,3),(4,0),(4,2) …
An improved method:
Consider the binomial expansion of (1−y)n=∑r=0n(nr).(−y)r
when y=1:(1−1)n=S1=∑r=0n(nr)(−y)r=0
when y=−1:(1+1)n=S2=∑r=0n(nr)(y)r=2n
Sum of even powers of y = SE=S1+S22
Sum of odd powers of y = SO=S1−S22
Applying this concept:
Substituting x=1 would give the sum of all terms
⟹S1=(2.12−3.1+1)11=0
⟹S2=(2.(−1)2−3.(−1)+1)11=611
Sum of even powers = SE=S1+S22=6112
Ans: 6112
What is sum of coefficients of all the even powers of x in the expansion of (2x^2 - 3x + 1)?
Sum of the last 30 coefficient s in the expansion of (1+x) ^59 when expanded in ascending powers of x?
The sum of the coefficients of odd powers of x in the expansion of (1-x+x²-x³) ^9 is?
What is the sum of coefficients in the expansion of (x + 2y + Z) ^10?
What is the coefficient of x^11 in the expansion of (1-2x+3x^2) (1+x) ^11?
What is the sum of coefficients of all the integral powers of x in the expansion of (1+2√x) ^40?
What is the coefficient of the term x^3 in the expansion of (2x-1) ^9?
Let P(x) = (2x^2−2x+1) ^17* (3x^2−3x+1) ^17. What is the sum of all coefficients at even powers of x?
What is the coefficient of x11 in the expansion of (1+3x+2x2)6 ?
What is the coefficient of x in the expansion of (1-2x³+3x^(5)) (1+(1/x)) ^(8)?
The sum of the coefficients of odd powers of x in the expansion of (1-x+x²-x³) ^9 is?
Let
(1−x+x2−x3)9=a0+a1x+a2x2+⋯+a26x26+a27x27 .
If you plug in x=1 , then you get
0=a0+a1+a2+⋯+a26+a27 . (1)
If you plug in x=−1 , then you get
49=a0−a1+a2−⋯+a26−a27 . (2)
Now take equation (1) and subtract equation (2) to get
−49=2a1+2a3+⋯+2a25+2a27 .
Finally, just divide both sides by 2 to get
−492=a1+a3+⋯+a25+a27 .
So, the sum of the coefficients of the odd powers of x is −492=−131072 .
What is the sum of coefficients in the expansion of (x + 2y + Z) ^10?
2^20 is the answer
Trick: when ever you come across problems of this type just put “1″ as the value of variables in the question
Ex: in the above question put x=y=z=1 and
We get (1+2+1)^10
4^10
Therefore 2^20 is the sum of co efficients.
What is sum of coefficients of all the even powers of x in the expansion of (2x^2 - 3x + 1)?
I wonder whether the question is quite correctly stated.
The answer as it stands is simply 2+1=3 . What else?
I suspect the question should have a natural-number power in it:
What is the sum of the coefficients of all the even powers of x in (2x2−3x+1)n ?
Consider the results of evaluating this for both x=1 and x=−1 . The sum must be twice the sum of the even-power terms, and the difference twice the sum of the odd-power terms.
The answer comes out, therefore, as 12(0n+6n) or simply 126n . For n=1 this is 3 , as noted before.