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Triangle numbers

Source: ATM Mathematics Teaching 239

Let

T_{n}

be the

n^{th}

triangle number. Find

n

such that:

T_{n} + T_{n + 1} + T_{n + 2} + T_{n + 3} = T_{n + 4} + T_{n + 5}

Show answer & extension

Hide answer & extension

T_{n} = \frac{1}{2} n (n + 1)

, so:

T_{n} + T_{n + 1} = \frac{1}{2} n (n + 1) + \frac{1}{2} (n + 1) (n + 2)

= (n + 1)^{2}

So, we are looking for

n

such that

(n + 1)^{2} + (n + 3)^{2} = (n + 5)^{2}

. This is true when

n = 5

(

6^{2} + 8^{2} = 10^{2}

).

Extension

Find

n

such that

T_{n} + T_{n + 1} + T_{n + 1} + T_{n + 2} = T_{n + 2} + T_{n + 3}

.

Tags: numbers, triangle numbers

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