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3 years ago

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Consider two investments, one earning simple interest and one earning compound interest. If both start with the same initial de

posit (and you make no other deposits or withdrawals) and earn the same annual interest rate, how will the balance in the simple interest account compare to that of the compound interest?

1 answer:

Len [333]3 years ago

6 0

Answer:

Ratio between balances will be:

$(x*(1+i)*n)/(x*(1+i)^n)$

Where;

x = deposit

%i = interest rate annual

n = years

Step-by-step explanation:

Lets say first deposit is x$ for both investment and annual interest rate for both investment are %i. Also, they stayed under i interest in n years:

Total balance for simple interest is:

$Total balance=x*(1+i)*n$

How ever total balance for compound interest is:

$Total balance=x(1+i)^n$

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The following are the solution to the given points:

Step-by-step explanation:

Given value:

$1) \sum ^{\infty}_{k = 1} \frac{1}{k+1} - \frac{1}{k+2}\\\\2) \sum ^{\infty}_{k = 1} \frac{1}{(k+6)(k+7)}$

Solve point 1 that is $\sum ^{\infty}_{k = 1} \frac{1}{k+1} - \frac{1}{k+2}\\\\$ :

when,

$k= 1 \to s_1 = \frac{1}{1+1} - \frac{1}{1+2}\\\\$

$= \frac{1}{2} - \frac{1}{3}\\\\$

$k= 2 \to s_2 = \frac{1}{2+1} - \frac{1}{2+2}\\\\$

$= \frac{1}{3} - \frac{1}{4}\\\\$

$k= 3 \to s_3 = \frac{1}{3+1} - \frac{1}{3+2}\\\\$

$= \frac{1}{4} - \frac{1}{5}\\\\$

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Calculate the sum $(S=s_1+s_2+s_3+......+s_n)$

$S=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.....\frac{1}{n+1}-\frac{1}{n+2}\\\\$

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$\boxed{\text{In point 1:} \sum ^{\infty}_{k = 1} \frac{1}{k+1} - \frac{1}{k+2} =\frac{4}{5}}$

In point 2: $\sum ^{\infty}_{k = 1} \frac{1}{(k+6)(k+7)}$

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$k= 1 \to s_1 = \frac{1}{(1+6)(1+7)}\\\\$

$= \frac{1}{7 \times 8}\\\\= \frac{1}{56}$

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$S= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}....+\frac{1}{(n+6)(n+7)}\\\\$

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$\boxed{\text{In point 2:} \sum ^{\infty}_{k = 1} \frac{1}{(n+6)(n+7)} = 0.066}$

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<em>Comment on factoring</em>

There are a number of ways to solve quadratics. Apart from using a graphing calculator, one of the easiest is factoring. Here, we're looking for factors of -486 that have a sum of 9.

486 = 2 × 3^5, so we might guess that the factors of interest are -2·3² = -18 and 3·3² = 27. These turn out to be correct: -18 +27 = 9; (-18)(27) = -486.

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