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2008th Term of the Sequence 1, 2,2,2, 3,3,... | JEE

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

Question
The 20082008th term of the sequence 1, 2,2,2, 3,3,3,3,3,3, 4,…1,\ 2,2,2,\ 3,3,3,3,3,3,\ 4,\ldots where nn occurs n(n+1)2\frac{n(n+1)}{2} times, equals
A2424
B2323
C2222correct
D2121
Solution
Step 1: Group rr has length r(r+1)2\dfrac{r(r+1)}{2}, so the cumulative count up to group rr is
∑j=1rj(j+1)2=r(r+1)(r+2)6.\sum_{j=1}^{r}\frac{j(j+1)}{2}=\frac{r(r+1)(r+2)}{6}.
 Step 2: If the 20082008th term is in group rr, then
(r−1)r(r+1)6<2008≤r(r+1)(r+2)6.\frac{(r-1)r(r+1)}{6}<2008\le\frac{r(r+1)(r+2)}{6}.
 Step 3: Group 2121 ends at 21⋅22⋅236=1771\dfrac{21\cdot 22\cdot 23}{6}=1771 and group 2222 at 22⋅23⋅246=2024\dfrac{22\cdot 23\cdot 24}{6}=2024. Since 1771<2008≤20241771<2008\le 2024, the term is 22.22. Correct answer: (3)
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