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$\begingroup$

Find $A$:

$$A=+\frac{3}{4×8}-\frac{3×5}{4×8×12}+\frac{3×5×7}{4×8×12×16}-···$$

My Try :

$$a_1=\frac{3}{4×8}-\frac{3×5}{4×8×12}=\frac{3×12-3×5}{4×8×12}=\frac{3(12-5)}{4×8×12}=\frac{3(7)}{4×8×12}$$

$$a_2=\frac{3(7)}{4×8×12}-\frac{3×5×7}{4×8×12×16}=\frac{3×7×16-3×5×7}{4×8×12×16}=\frac{3×7(16-7)}{4×8×12×16}\\=\frac{3×7(8)}{4×8×12×16}$$

now?

  • 0
    What is the order of signs? $+, -, -, +$ repeated? – 2017-02-24
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    @didgognsexcuse me . Edited by me – 2017-02-24
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    Answer is $\sqrt{\frac 23}-\frac 34$ – 2017-02-24

1 Answers 1

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$$A=\sum_{n=2}^\infty (-1)^{n} \frac {(2n)!/(2^n n!)}{4^n n!}$$ $$=\sum_{n=2}^\infty (-1)^{n} \frac {(2n)!}{8^{n} (n!)^2}$$ $$=\sum_{n=2}^\infty \left(-\frac{1} {8}\right)^{n} \binom {2n} {n}$$

Now, note that $$\frac{1}{\sqrt{1-4x}} = \sum_{n=0}^\infty \binom {2n} {n}x^n$$ and take $x=-1/8$, separate 1st and 2nd term of the infinite sum to find your answer.

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    Not quite. $C_n = \frac{1}{n+1}\binom{2n}{n}$ – 2017-02-24
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    Uh oh. I will fix it. – 2017-02-24