Why do I need a $1$ in this equation of growth and decay? $A=P(1\pm r)^t$
Why must I divide by $2$ to find the sum of an arithmetic sequence?
$S(n) = \frac{n}{2}(a_1 + a_n)$
Why do I need a $1$ in this equation of growth and decay? $A=P(1\pm r)^t$
Why must I divide by $2$ to find the sum of an arithmetic sequence?
$S(n) = \frac{n}{2}(a_1 + a_n)$
On
1 - You don't really need the 1 in the exponential formula. any exponential function can be written in the form $$f(x)=a(b^x) $$ where $b$ is the common ratio It is good to understand that ...
if $b=1$ there is no growth or decay
if $b>1$ then $|f|$ is increasing by a growth factor of $(b-1)$
if $b<1$ then $|f|$ is decreasing by a decay factor of $(1-b)$
2 - The arithmetic sum is based on the observation that if .. $$t_n = a + (n-1)d $$ then for every $k$ satisfying $1\le k\le n$ $$ t_k+t_{n+1-k} = 2a + [(k-1) + (n+1-k-1)]d \\= 2a+(n-1)d = t_1+t_n$$
which is independent of $k$ so we can conclude that the average value of each term in the series in $\frac{t_1+t_n}2$.
Since the series contains $n$ terms, the sum will be ... $$S_n = n\left( \frac{t_1+t_n}2 \right )$$
On
1)You need to have the one there because if you have exponential growth it means that you are adding on to the initial value. Example problem: If I earn 10% every year with \$1000 every year then how much will I have in 2 years. Answer: I am earning 10% so the equation is: A = 1000(1+0.1)^2 This means that I would get a total of \$1210. If we did not add the one then the equation would be: A = 1000(0.1)^2 This means that I would get $10 which means that I lost money.
2)For your second question, I have a simple way of thinking about it. Example: 0+1+2+3+4+5+6+7+8+9 = 9+9+9+9+9 = 45 Add the first and the last term and you get 9. Do the same for the second and ninth and you get 9. Do the same for the fifth and the sixth terms and you get 9. So all you got to do is for all the terms and you divide by two so that you don't get repeats.
1) because for $t=0$ we have $P=A$
2)because in the arithmetic sequence the addition of two terms equidistant from the extreme elements $a_1$ and $a_n$ is the same, and we have $n/2$ sums of this kind.