In the polynomial 3x^4−3x^3+6x^2−3x+6, which term would represent the hundreds place, if this were an integer?

Mathematics

1 answer:

Over [174]4 years ago

5 0

Answer:

−3x^3

Step-by-step explanation:

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Please help me out on this !!!

lawyer [7]

Answer:

Tu is your answer hope it helps you

8 0

3 years ago

Determine whether the series is absolutely convergent 1-(1*3/3! (1*3*5/5!-（3*5*7）

DENIUS [597]

It's not clear what your series is, so I'm going to take a wild guess on what it is you mean:

$1-\dfrac{1\times3}{3!}+\dfrac{1\times3\times5}{5!}-\dfrac{1\times3\times5\times7}{7!}+\cdots$
$=\displaystyle\sum_{n=1}^\infty\frac{(-1)^{n-1}}{(2n-1)!}\prod_{k=1}^n(2k-1)$

For the sum to be absolute convergent, the sum of the absolute value of the summand must converge, so you are really examining the convergence of

$\displaystyle\sum_{n=1}^\infty\frac1{(2n-1)!}\prod_{k=1}^n(2k-1)$

This is easily checked with the ratio test:

$\displaystyle\lim_{n\to\infty}\left|\frac{\displaystyle\dfrac1{(2(n+1)-1)!}\prod_{k=1}^{n+1}(2k-1)}{\displaystyle\dfrac1{(2n-1)!}\prod_{k=1}^n(2k-1)}\right|=\lim_{n\to\infty}\left|\frac{\dfrac{1\times3\times5\times\cdots\times(2n-1)\times(2n+1)}{(2n+1)!}}{\dfrac{1\times3\times5\times\cdots\times(2n-1)}{(2n-1)!}}\right|$
$\displaystyle=\lim_{n\to\infty}\left|\frac{\dfrac{2n+1}{(2n+1)(2n)}}{\dfrac11}\right|=\lim_{n\to\infty}\dfrac1{2n}=0$

Since $\sum|a_n|$ converges by the ratio test, the series $\sum a_n$ converges absolutely.