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Therefore, the condition for the constant term is: n − 2k = 0 ⇒ k = n 2 . In other words, in this case, the constant term is the middle one (k = n 2). Case 3: If the terms of the binomial are two distinct variables x and y, such that y cannot be expressed as a ratio of x, then there is no constant term . This is the general case (x +y)n ...
The constant term in a binomial expansion is the one, if it exists, that has no variable terms within it. This prior answer on Socratic has a great explanation on how to find the constant term: How do I use the binomial theorem to find the constant term? Following the instructions in that answer, we first set up the general term:
Therefore, the condition for the constant term is: #n-2k=0 rArr# #k=n/2#. In other words, in this case, the constant term is the middle one (#k=n/2#). Case 3: If the terms of the binomial are two distinct variables #x# and #y#, such that #y# cannot be expressed as a ratio of #x#, then there is no constant term. This is the general case #(x+y)^n#
Help. Constant term is -5376 (r+1)^ (th) term in the expansion of (a+b)^n is C_r^na^ (n-r)b^r Hence (r+1)^ (th) term in the expansion of (2x-1/x^2)^9 is C_r^9 (2x)^ (9-r) (-1/x^2)^r = C_r^9*2^ (9-r)*x^ (9-r) (-1)^r/x^ (2r) = C_r^9*2^ (9-r)*x^ (9-r-2r) (-1)^r = C_r^9*2^ (9-r) (-1)^r*x^ (9-3r) As we are seeking constant term, this means power of ...
Now for this term to be the constant term, x^ (3-r) should be equal to 1. Therefore, x^ (3-r) = x^0. => 3-r =0. => r=3. Thus, the fourth term in the expansion is the constant term. By putting r=3 in the general term, we will get the value of the constant term. Answer link. Let (2x+3) ^3 be a given binomial. From the binomial expression, write ...
Any term that doesn't have a variable in it is called a "constant" term. types of polynomials depends on the degree of the polynomial. x5 = quintic. x4 = quadratic. x3 = cubic. x2 = quadratic. Answer link. degree= 0 type= constant leading coefficient= 0 constant term= -6 -6 is the product of this equation therefore there are no constant term or ...
For the Const. Term, the index of x must be 0. ∴ 12 −3r = 0 ⇒ r = 4. (∗) ⇒ T 4+1 = T 5 = 6C4(2)6−4(− 1)4 x12−(3) (4), = 6C2 ⋅ 22 ⋅ (1), = (6)(5) (1)(2) ⋅ 4. Hence, the desired const. term is 60, and is the 5th term in. the Expansion. Enjoy Maths.!
Degree: 8 Leading term: 3x^5 Leading Coefficient: 3 Constant: 1 End behavior: See below in blue The degree is the sum of the exponents on all terms. Our exponents are 5, 2 and 1, which sum up to 8. This is the degree of our polynomial g(x). The leading term of a polynomial is just the term with the highest degree, and we see this is 3x^5. The leading coefficient is just the number multiplying ...
If the constant term of the binomial expansion (2x− 1 x)n is −160, then n is equal to -. View Solution. Click here:point_up_2:to get an answer to your question :writing_hand:how do i find the constant term of a binomial expansion.
The leading coefficient is the coefficient of the hightest exponent of the variable. The constant term is the term not multiplied by the variable. See explanation. The degree of this polynomial is 5, the leading coefficient is -2 and the constant term is 9. The degree of a polynomial is the highest exponent with a non-zero coefficient.