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The sum of $$(1\times4)+(2\times7)+(3\times10)+(4\times13)+\cdots$$ for $$49$$ terms is equal to
The $$n$$th term is $$n(3n+1)=3n^2+n$$. Therefore the sum is $$3\sum_{n=1}^{49}n^2+\sum_{n=1}^{49}n$$. Using the standard sums gives $$3\cdot\frac{49\cdot50\cdot99}{6}+\frac{49\cdot50}{2}=122500$$.
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