Find the sum of the series up to $90$ terms: $1+2+3+3+4+6+5+6+9+ $ ....

Regrouping the series into sets of three terms, we get $(1+2+3) + (3+4+6) + (5+6+9) + ...$, where the sums of the groups are 6, 13, and 20 respectively. These sums form an arithmetic progression with a first term of 6 and a common difference of 7. Using the sum formula for an arithmetic progression, the sum of the first 30 groups (which equals 90 terms) is $30/2 \times [2(6) + (30-1)7] = 15 \times 215 = 3225$. The result is 3225.

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