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# Algebra - Algebraic Manipulation Problem

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For a positive integer n , the  nth triangular number is $$T(n)=\dfrac{n(n+1)}{2}.$$

For example, $$T(3) = \frac{3(3+1)}{2}= \frac{3(4)}{2}=6$$ , so the third triangular number is 6.

Determine the smallest integer $$b>2011$$  such that $$T(b+1)-T(b)=T(x)$$ for some positive integer x .

Aug 2, 2022

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Aug 2, 2022
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$$\frac{(b+1)(b+2)}{2}-\frac{b(b+1)}{2}=\frac{x(x+1)}{2}$$

$$2(b+1)=x(x+1)$$

$$\frac{x(x+1)}{2}-1=b>2011$$

So:

$$x(x+1)>4024$$

if x=63, then this satisfies the condition, and is the smallest positive integer x. (As x=62 won't be greater than 4024).

So:

$$2(b+1)=4032 \implies b=2015$$ satisfies the condition.

Check:

$$T(2016)-T(2015)=2033136-2031120=2016=\frac{63(63+1)}{2}=T(63)$$

Aug 2, 2022
#3
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$$T(b+1) = \frac{(b+1)(b+2)}{2}=\frac{b^2+3b+2}{2}$$

$$T(b) = \frac{b(b+1)}{2}=\frac{b^2+b}{2}$$

$$T(b+1)-T(b)=b+1$$

$$b+1$$ should be a triangular number.

The smallest triangular number after 2011 is 2016 (take 2011, double it, square root the result, round that, plug it in as n. It will at the very least get you close to the triangular number wanted and then you will have know that number is the nth triangular number, making finding the next easy).

$$b+1=2016$$

$$b=2015$$

Aug 2, 2022