document.write( "Question 815543: Hi i need help on this question, please help:
\n" ); document.write( "Carol and Anthony will invest £5100 per year in a savings account at a rate of 4% per year compounded annually. How much will the JOINT total be after 5 years.\r
\n" ); document.write( "\n" ); document.write( "Can someone please explain how to work it out or what formula to use.
\n" ); document.write( "Thank you!
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Algebra.Com's Answer #491671 by KMST(5328)\"\" \"About 
You can put this solution on YOUR website!
At the start, they deposit £5100.
\n" ); document.write( "In the first year the interest earned is the initial balance at the start of that year (£5100) times the interest rate as a fraction or decimal
\n" ); document.write( "\"%22%28+4+%25+=%22+\"\"4%2F100=0.04\" .
\n" ); document.write( "At the end of the first year, they will have the initial amount, plus the interest:5100
\n" ); document.write( "\"%22%A3+5100+%2B+%A3%22\"\"5100%2A0.04=+%22%A3%22\"\"%281%2B0.04%29=+%22%A3%22\"\"5100%2A1.04\" .
\n" ); document.write( "That means that in the first year the initial balance grows by a factor of \"1.04\" .
\n" ); document.write( "That happens no matter what the initial balance, and no matter what happened before.
\n" ); document.write( " At the beginning of the second year, they deposit £5100 again,
\n" ); document.write( "and end up with (in £)
\n" ); document.write( "\"5100%2A1.04%2B5100\"
\n" ); document.write( "Over the second year that amount of money again multiplies times \"1.04\" to get to
\n" ); document.write( "\"%285100%2A1.04%2B5100%291.04=5100%2A1.04%5E2%2B5100%2A1.04\"
\n" ); document.write( "by the end of the second year.
\n" ); document.write( "
\n" ); document.write( "At the beginning of the third year, they deposit £5100 again,
\n" ); document.write( "and end up with (in £)
\n" ); document.write( "\"5100%2A1.04%5E2%2B5100%2A1.04%2B5100\"
\n" ); document.write( "Over the third year that amount of money again multiplies times \"1.04\" to get to
\n" ); document.write( "
\n" ); document.write( "by the end of the third year.
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\n" ); document.write( "The pattern continues, and by the end of the fifth year,their balance is
\n" ); document.write( "\"5100%2A1.04%5E5%2B5100%2A1.04%5E4%2B5100%2A1.04%5E3%2B5100%2A1.04%5E2%2B5100%2A1.04\"
\n" ); document.write( "=\"5100%2A1.04%2A%281.04%5E4%2B1.04%5E3%2B1.04%5E2%2B1.04%2B1%29\"
\n" ); document.write( "=\"5100%2A1.04%2A%281.04%5E5-1%29%2F%281.04-1%29=5100%2A1.04%2A%281.04%5E5-1%29%2F0.04\"
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