方法一为常规法,方法二为大聪明观察法,大家自行选择。 例题二:Find the term independent ofxin the expansion of(4x^3+\frac{1}{2x})^8 首先解释下这道题中出现的常见表达方式 term independent of x,表面意思是独立于x的项,实际意思是常数项(constant term),换个更有用的说法,我们需要找到x^0项。 方法...
(a) For (x^(2)-(1)/(x))^(n) T(r+1)=^(n)C(r )(x^(2))^(n-r)(-1)^(r )x^(-r)=^(n)C(r )x^(2n-3r)(-1)^(r ) Constant term =^(n)C(r )(-1)^(r ) if 2n=3r i.e. coefficient of x=0 Hence .^(n)C(2n//3)(-1)^(2n//3)=15=^(6)C(4)...
How to find the term independent in x or constant term in a binomial expansion, examples and step by step solutions, Binomial Expansion with fractional powers or powers unknown, A Level Maths
2.a taxonomic name consisting of a generic and a specific term, used to designate species. adj. 3.of or pertaining to a term, expression, or quantity that has two parts. [1550–60; < Late Latinbinōmi(us)having two names] bi•no′mi•al•ism,n. ...
The binomial theorem states that in the expansion of (x + a)n, the coefficients are the combinatorial numbers nCk , where k—the exponent of a—successively takes the values 0, 1, 2, . . . , n.Each term in the expansion will have this form: n (n − k) k an − kbk ....
We now introduce the binomial expansion of (1-t2)-1/2 and integrate term-by-term. The result is Eq. (2.63). ▪ Symbolic Computation of Binomial Coefficients Both maple and mathematica support procedures that compute binomial coefficients. We have nm:binomial(n,m)(MAPLE),Binomial[n,m](...
of each term is a positive integer and the value depends on ‘n’ and ‘b’. for example, for n=4, the expansion (x + y) 4 can be expressed as (x + y) 4 = x 4 + 4x 3 y + 6x 2 y 2 + 4xy 3 + y 4 the coefficients of the binomials in this expansion 1,4...
I assumed that (nCr) is not a constant, as I expect, it must be a function of (n and r). If it was a constant of (n and r), then define it outside the loop.
∴ 4th term is the one independent of x, i.e., the constant term. Try these questions Expand (x + 1/y)7 Find the middle term of the following expansion (x/a + y/b)6 (√a – b)8 (x2/y – y2/x)8 (xy – 1/x2y2)5 (a/x + x/a)5 (x – 3/y)5 Find the term...
If (n+1) is odd, then clearly there is only one middle term, viz., the (n2+1)(n2+1) th term tn2+1tn2+1 in the expansion of (a+b)n;(a+b)n; but if (n+1)(n+1) is even, then obviously there are two middle terms, viz., the n+12thn+12th term and the (n+12+1)...