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3 years ago

Find the first, fourth, and eighth terms of the sequence. A(n) = –5 • 3n–1

Mathematics

2 answers:

jeka943 years ago

6 0

Answer:

Step-by-step explanation:

$\text{Substitute}\ n=1,\ n=4\ \text{and}\ n=8\ \text{to}\ A(n)=-5\cdot3^{n-1}:\\\\A(1)=-5\cdot3^{1-1}=-5\cdot3^0=-5\cdot1=-5\\\\A(4)=-5\cdot3^{4-1}=-5\cdot3^3=-5\cdot27=-135\\\\A(8)=-5\cdot3^{8-1}=-5\cdot3^7=-5\cdot2,187=-10,935$

Sindrei [870]3 years ago

3 0

Answer:

Step-by-step explanation:

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Find the missing value.-7/13 = m/104

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Evaluate x — + 6(x-12) when x=12 4

Ghella [55]

Answer:

Step-by-step explanation:

x=12

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3 years ago

For the following telescoping series, find a formula for the nth term of the sequence of partial sums

gtnhenbr [62]

I'm guessing the sum is supposed to be

$\displaystyle\sum_{k=1}^\infty\frac{10}{(5k-1)(5k+4)}$

Split the summand into partial fractions:

$\dfrac1{(5k-1)(5k+4)}=\dfrac a{5k-1}+\dfrac b{5k+4}$

$1=a(5k+4)+b(5k-1)$

If $k=-\frac45$ , then

$1=b(-4-1)\implies b=-\frac15$

If $k=\frac15$ , then

$1=a(1+4)\implies a=\frac15$

This means

$\dfrac{10}{(5k-1)(5k+4)}=\dfrac2{5k-1}-\dfrac2{5k+4}$

Consider the $n$ th partial sum of the series:

$S_n=2\left(\dfrac14-\dfrac19\right)+2\left(\dfrac19-\dfrac1{14}\right)+2\left(\dfrac1{14}-\dfrac1{19}\right)+\cdots+2\left(\dfrac1{5n-1}-\dfrac1{5n+4}\right)$