Sunday Afternoon Maths XXVII

Hide answer & extension

Hide answer & extension

$$n^2 \leq T_a$$ $$n^2 \leq \frac{1}{2}a(a+1)$$ $$2n^2 \leq a^2+a$$ $$2n^2 \leq 2a^2$$ $$n^2 \leq a^2$$ $$n \leq a$$ $$T_{a+2}=\frac{1}{2}(a+2)(a+3)$$ $$=\frac{1}{2}(a^2+5a+6)$$ $$=\frac{1}{2}(a^2+a)+\frac{1}{2}(4a+6)$$ $$=\frac{1}{2}a(a+1)+2a+3$$ $$=T_a+2a+3$$ $$T_a+2a+3 \geq n^2+2n+3$$ $$>n^2+2n+1 = (n+1)^2$$

Extension