To write the expression \( 4a + 1b + 2a + 2b \) using repeated addition for each group, we can break it down as follows:
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4 groups of \( a \): \[ a + a + a + a \]
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1 group of \( b \): \[ b \]
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2 groups of \( a \): \[ a + a \]
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2 groups of \( b \): \[ b + b \]
Now, we can put it all together:
\[ (a + a + a + a) + (b) + (a + a) + (b + b) \]
Finally, to write the equivalent expression with the fewest possible terms, we combine like terms:
\[ (4a + 2a) + (1b + 2b) \]
This simplifies to:
\[ 6a + 3b \]
So the final equivalent expression with the fewest possible terms is:
\[ 6a + 3b \]