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JEE Main 28-Jan-2025 Paper - Shift 2 — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEEJEE Main 28-Jan-2025 Paper - Shift 24 Marks
MCQ
For positive integers $n$, if $4 a_{n}=\left(n^{2}+5 n+6\right)$and$\mathrm{S}_{\mathrm{n}}=\sum_{\mathrm{k}=1}^{\mathrm{n}}\left(\frac{1}{\mathrm{a}_{\mathrm{k}}}\right)$, then the value of $507 \mathrm{~S}_{2025}$ is :
  • A
    540
  • B
    1350
  • C
    675
  • D
    135
✓

Answer

C.
$\mathrm{a}_{\mathrm{n}}=\frac{\mathrm{n}^{2}+5\mathrm{n}+6}{4}$
$\begin{aligned}\mathrm{S}_{\mathrm{n}} & =\mathrm{S}_{\mathrm{n}}=\sum_{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{\mathrm{a}_{\mathrm{k}}}=\sum_{1}^{\mathrm{n}} \frac{4}{\mathrm{k}^{2}+5 \mathrm{k}+6} \\& =4 \sum_{\mathrm{k}=1}^{\mathrm{n}} \frac{1}{(\mathrm{k}+2)(\mathrm{k}+3)} \\& =4 \sum_{\mathrm{k}=1}^{\mathrm{n}} \frac{1}{\mathrm{k}+2}-\frac{1}{\mathrm{k}+3} \\& =4\left(\frac{1}{3}-\frac{1}{4}\right)+4\left(\frac{1}{4}-\frac{1}{5}\right)+\ldots \ldots . .\end{aligned}$
$\begin{gathered}4\left(\frac{1}{\mathrm{n}+2}-\frac{1}{\mathrm{n}+3}\right) \\=4\left(\frac{1}{3}-\frac{1}{\mathrm{n}+3}\right) \\=\frac{4 \mathrm{n}}{3(\mathrm{n}+3)} \\{ }^{507} \mathrm{~S}_{2025}=\frac{(507)(4)(2025)}{3(2028)}\end{gathered}$
$=675$

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