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Mathematics

1 answer:

algol133 years ago

6 0

Answer:

B (6, 8)

C (1, 5)

see attached diagram (qu 1 is blue, qu2 is red)

Step-by-step explanation:

A = (2, 3)

If B is 4 right and 5 up from point A, then

B = (2 + 4, 3 + 5) = (6, 8)

2. C = (9, 7)

If D is 8 left and 2 down from point C, then

C = (9 - 8, 7 - 2) = (1, 5)

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Find the sum of series 4/(4n-3) (4n+1)

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$\dfrac4{(4n-3)(4n+1)}=\dfrac1{4n-3}-\dfrac1{4n-1}$

Assuming the sum starts at $n=1$ , the $N$ th partial sum is

$\displaystyle\sum_{n=1}^N\left(\frac1{4n-3}-\frac1{4n-1}\right)=\left(1-\frac15\right)+\left(\frac15-\frac19\right)+\cdots+\left(\frac1{4N-7}-\frac1{4N-3}\right)+\left(\frac1{4N-3}-\frac1{4N+1}\right)$
$\displaystyle\sum_{n=1}^N\left(\frac1{4n-3}-\frac1{4n-1}\right)=1-\frac1{4N-1}$

As $N\to\infty$ , you're left with simply 1.