find in ascending powers of x the first three terms of (2+x)(1+x)^8
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Answer:
How do I find the first three terms, in descending powers of X, in the expansion of (x- 2/X) ^5?
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How do I find the first three terms, in descending powers of X, in the expansion of (x- 2/X) ^5?
From the binomial theorem,
(a+b)n=∑nk=0(nk)an−kbk
=(n0)anb0+(n1)an−1b1+(n2)an−2b2+…
=an+nan−1b+n(n−1)2an−2b2+…
We’ll now replace n by 5 , a by x and b by −2X , resulting in:
x5+5x4×−2X+10x3×(−2)2X2+…
=x5−10x4X+40x3X2+…
=x5−10x4X−1+40x3X−2+…
That’s the first three terms in decreasing powers of X
In the above, I've assumed that x and X are different variables. If, however, x=X , then the answer becomes:
X5−10X3+40X+…
Answer:
From the binomial theorem,
(a+b)n=∑nk=0(nk)an−kbk
=(n0)anb0+(n1)an−1b1+(n2)an−2b2+…
=an+nan−1b+n(n−1)2an−2b2+…
We’ll now replace n by 5 , a by x and b by −2X , resulting in:
x5+5x4×−2X+10x3×(−2)2X2+…
=x5−10x4X+40x3X2+…
=x5−10x4X−1+40x3X−2+…
That’s the first three terms in decreasing powers of X
In the above, I've assumed that x and X are different variables. If, however, x=X , then the answer becomes:
X5−10X3+40X+…