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Question 22

The numbers $$1,3,6,10,\ldots$$ are called triangular numbers. The $$n$$th triangular number is $$T_n=\frac{n(n+1)}{2}$$. Then the value of $$T_{3n+1}-9T_n$$ is equal to


Correct Answer: 1

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We have $$T_{3n+1}=\dfrac{(3n+1)(3n+2)}{2}$$ and $$9T_n=\dfrac{9n(n+1)}{2}$$. 

$$T_{3n+1}-9T_n =\dfrac{(3n+1)(3n+2)}{2}-\dfrac{9n(n+1)}{2} = \dfrac{(9n^2+9n+2)-9n^2-9n}{2} =1$$

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