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Largest Term of the Sequence n^2/(500+3n^3) |

JEE Maths question with a full step-by-step solution.

Question
The largest term of the sequence 1503,4524,9581,16692,…\frac{1}{503},\frac{4}{524},\frac{9}{581},\frac{16}{692},\ldots is
A16692\frac{16}{692}
B4524\frac{4}{524}
C491529\frac{49}{1529}correct
D135\frac{1}{35}
Solution
Step 1: The numerators are n2n^2 and the denominators are 500+3n3500+3n^3, so
Tn=n2500+3n3.T_n=\frac{n^2}{500+3n^3}.
 Step 2: Set dTndn=0\dfrac{dT_n}{dn}=0:
2n(500+3n3)−n2(9n2)(500+3n3)2=0 ⇒ 1000−3n3=0 ⇒ n=(10003)1/3∈(6,7).\frac{2n(500+3n^3)-n^2(9n^2)}{(500+3n^3)^2}=0\ \Rightarrow\ 1000-3n^3=0\ \Rightarrow\ n=\left(\frac{1000}{3}\right)^{1/3}\in(6,7).
 Step 3: The term is largest at n=7n=7:
T7=49500+3⋅343=491529.T_7=\frac{49}{500+3\cdot 343}=\frac{49}{1529}.
 Correct answer: (3)
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