Question

HELP ASAP
A high-interest savings account pays 5.5% interest compounded annually. If $300 is deposited initially and again at the first of each year, which summation represents the money in the account 10 years after the initial deposit?

Answer 1

Answer: $\sum_{n=1}^{10} 316.5(1.055)^{n-1}$

Step-by-step explanation:

If $ 300 is deposited in an account which pays 5.5% interest compounded annually.

Then the amount after 1 year,

$A = 300(1+\frac{5.5}{100}) = 300(1+0.055) = 3000(1.055)=316.5$

Similarly, In the second year the amount in his account,

= $316.5 + 316.5(1.055)$

In third year the amount in his account =  $316.5 +316.5(1.055)+ 316(1.055)^2$

In fourth year = $316.5 + 316.5(1.055) + 316(1.055)^2 + 316.5(1.055)^3$

According to the question,

He repeat this process up to 10 years,

Hence, in the 10 years the amount in his account

= $316.5 + 316.5(1.055) + 316(1.055)^2 + 316.5(1.055)^3$ --------- up to 10 terms,

= $\sum_{n=1}^{10} 316.5(1.055)^{n-1}$

Answer 2

The money in the account can be obtained by compound interest for formula. In the given problem we have interest rate annually, so the 'compound Interest when Interest is compounded yearly' formula is applicable.

The correct option is $\sum_{n=1}^{10}316.5(1.055)^{n-1}$

Given:

Annually interest rate $R=5.5\%$

Principal amount $P=\ 300$

Deposit time $n=10$.

Write the compound interest formula.

$A = P\left (1 + \dfrac {R}{100}\right)^n$

Substitute the value in above expression to find the compound interest 1 year.

$A=300\left (1 + \dfrac {5.5}{100}\right)^{1}\\A=300\times ( 1.055)$

Similarly, find the compound interest for 2 year.

$A=316.5+316.5(1.055)$

From the given question this process repeat for 10 year.

$A=316.5+316.5(1.055)+316.5(1.055)^2...$

The above expression can be written in form of summation.

$\sum_{n=1}^{10}316.5(1.055)^{n-1}$

Thus, the correct option is  $\sum_{n=1}^{10}316.5(1.055)^{n-1}$

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