Let $$t_{1},t_{2}$$,... be real numbers such that $$t_{1}+t_{2}+…+t_{n} = 2n^{2}+9n+13$$, for every positive integer $$n \geq 2$$. If $$t_{k}=103$$, then k equals
Correct Answer: 24
It is given that $$t_{1}+t_{2}+…+t_{n} = 2n^{2}+9n+13$$, for every positive integer $$n \geq 2$$.
We can say that $$t_{1}+t_{2}+…+t_{k} = 2k^{2}+9k+13$$ ... (1)
Replacing k by (k-1) we can say that
$$t_{1}+t_{2}+…+t_{k-1} = 2(k-1)^{2}+9(k-1)+13$$ ... (2)
On subtracting equation (2) from equation (1)
$$\Rightarrow$$ $$t_{k} = 2k^{2}+9k+13 - 2(k-1)^{2}+9(k-1)+13$$
$$\Rightarrow$$ $$103 = 4k+7$$
$$\Rightarrow$$ $$k = 24$$
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