Question:
Let t1, t2,… be real numbers such that t1+t2+…+tn = 2n2+9n+13, for every positive integer n ≥ 2. If tk=103, then k equals
Correct Answer: 24
$t_{1}+t_{2}+\ldots+t_{n}=2 n^{2}+9 n+13 \rightarrow(1)$
$t_{1}+t_{2}+\ldots+\ {tn}-1=2(n-1)^{2}+9(n-1)+13 \rightarrow(2)$
From $(2)-(1),$ we get $t_{n}=\left(2 n^{2}+9 n+13\right)-\left(2(n-1)^{2}\right.$
$+9(n-1)+13 )=4 n+7$
Given $t_{k}=103=>4 k+7=103 \Rightarrow k=24 \quad$
$t_{1}+t_{2}+\ldots+t_{n}=2 n^{2}+9 n+13 \rightarrow(1)$
$t_{1}+t_{2}+\ldots+\ {tn}-1=2(n-1)^{2}+9(n-1)+13 \rightarrow(2)$
From $(2)-(1),$ we get $t_{n}=\left(2 n^{2}+9 n+13\right)-\left(2(n-1)^{2}\right.$
$+9(n-1)+13 )=4 n+7$
Given $t_{k}=103=>4 k+7=103 \Rightarrow k=24 \quad$
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