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Let the positive integers be written in the form :
If the $k^{\text {th }}$ row contains exactly $k$ numbers for every natural number $k$, then the row in which the number 5310 will be, is _______
If the $k^{\text {th }}$ row contains exactly $k$ numbers for every natural number $k$, then the row in which the number 5310 will be, is _______
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The correct answer is:
103
$\begin{aligned} & \mathrm{S}=1+2+4+7+\ldots \ldots+\mathrm{T}_{\mathrm{n}} \\ & \mathrm{S}=1+2+4+\ldots \ldots \\ & \mathrm{Tn}=1+1+2+3+\ldots \ldots+\left(\mathrm{T}_{\mathrm{n}}-\mathrm{T}_{\mathrm{n}-1}\right) \\ & \mathrm{T}_{\mathrm{n}}=1+\left(\frac{\mathrm{n}-1}{2}\right)[2+(\mathrm{n}-2) \times 1] \\ & \mathrm{T}_{\mathrm{n}}=1+1+\frac{\mathrm{n}(\mathrm{n}-1)}{2} \\ & \mathrm{n}=100 \quad \mathrm{~T}_{\mathrm{n}}=1+\frac{100 \times 99}{2}=4950+1 \\ & \mathrm{n}=101 \quad \mathrm{~T}_{\mathrm{n}}=1+\frac{101 \times 100}{2}=5050+1=5051 \\ & \mathrm{n}=102 \quad \mathrm{~T}_{\mathrm{n}}=1+\frac{102 \times 101}{2}=5151+1=5152 \\ & \mathrm{n}=103 \quad \mathrm{~T}_{\mathrm{n}}=1+\frac{103 \times 102}{2}=5254 \\ & \mathrm{n}=104 \quad \mathrm{~T}_{\mathrm{n}}=1+\frac{104 \times 103}{2}=5357\end{aligned}$
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