Sequences & SeriesmediumPYQ · JEE Main · 2 Apr 2026 · Shift 2 (Afternoon)Free

Sum of Cubes over Odd Numbers Series (8 Terms) = 71 | JEE 2026

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

Question
The sum 131+13+231+3+13+23+331+3+5+⋯\dfrac{1^3}{1}+\dfrac{1^3+2^3}{1+3}+\dfrac{1^3+2^3+3^3}{1+3+5}+\cdots up to 8 terms, is
A7070
B7171correct
C7272
D7373
Solution
Step 1: The rr-th term has numerator 13+23+⋯+r3=(r(r+1)2)21^3+2^3+\cdots+r^3=\left(\dfrac{r(r+1)}{2}\right)^2 and denominator 1+3+5+⋯+(2r−1)=r21+3+5+\cdots+(2r-1)=r^2. So
Tr=(r(r+1)2)2r2=r2(r+1)2/4r2=(r+1)24=r2+2r+14.T_r=\frac{\left(\frac{r(r+1)}{2}\right)^2}{r^2}=\frac{r^2(r+1)^2/4}{r^2}=\frac{(r+1)^2}{4}=\frac{r^2+2r+1}{4}.
 Step 2: Sum the first nn terms:
Sn=∑r=1nTr=14∑r=1n(r2+2r+1)=14[n(n+1)(2n+1)6+2⋅n(n+1)2+n].S_n=\sum_{r=1}^{n}T_r=\frac14\sum_{r=1}^{n}\left(r^2+2r+1\right)=\frac14\left[\frac{n(n+1)(2n+1)}{6}+2\cdot\frac{n(n+1)}{2}+n\right].
 Step 3: Put n=8n=8:
S8=14[8⋅9⋅176+8⋅9+8]=14[204+72+8]=2844=71.S_8=\frac14\left[\frac{8\cdot 9\cdot 17}{6}+8\cdot 9+8\right]=\frac14\left[204+72+8\right]=\frac{284}{4}=71.
 Correct answer: (2)
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