Sequences & SeriesmediumPYQ · JEE Main · 6 Apr 2026 · Shift 1 (Morning)Free

Alternating Sum of Cubes = 1856 | JEE Main 2026

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

Question
The value of 13−23+33−⋯+1531^3-2^3+3^3-\cdots+15^3 is
A17061706
B18561856correct
C19821982
D24032403
Solution
Step 1: Let the series be 13−23+33−43+⋯+133−143+1531^3-2^3+3^3-4^3+\cdots+13^3-14^3+15^3. Group the first 14 terms as 7 pairs (n3−(n+1)3)\big(n^3-(n+1)^3\big), n=1,3,5,7,9,11,13n=1,3,5,7,9,11,13, plus 15315^3. Step 2: (n+1)3=n3+3n2+3n+1(n+1)^3=n^3+3n^2+3n+1:
n3−(n+1)3=−3n2−3n−1.n^3-(n+1)^3=-3n^2-3n-1.
 Step 3: Let Sp=∑n=1,3,…,13(−3n2−3n−1)S_p=\displaystyle\sum_{n=1,3,\ldots,13}(-3n^2-3n-1).
∑n=1+3+5+7+9+11+13=49,\sum n=1+3+5+7+9+11+13=49,
∑n2=1+9+25+49+81+121+169=455.\sum n^2=1+9+25+49+81+121+169=455.
⇒Sp=−3(455)−3(49)−7=−1365−147−7=−1519.\Rightarrow S_p=-3(455)-3(49)-7=-1365-147-7=-1519.
 Step 4: 153=337515^3=3375.
∴13−23+⋯+153=Sp+153=−1519+3375=1856.\therefore1^3-2^3+\cdots+15^3=S_p+15^3=-1519+3375=1856.
 Correct answer: (2)
Still stuck on this question?Ask your doubt on WhatsApp
Similar questions
Sequences & Series · easy
a,b,c,da, b, c, d are in increasing G.P. If the A.M. between aa and bb is 66 and the A.M. between cc and dd is 5454, then the A.M. of aa and dd is
Sequences & Series · easy
Let a1,a2,…a_1, a_2, \ldots and b1,b2,…b_1, b_2, \ldots be arithmetic progressions with a1=25a_1=25, b1=75b_1=75 and a100+b100=100a_{100}+b_{100}=100. Then the sum of the first hundred terms of a1+b1, a2+b2, …a_1+b_1,\ a_2+b_2,\ \ldots is
Sequences & Series · easy
If Sn=11⋅2⋅3+12⋅3⋅4+⋯+1n(n+1)(n+2)S_n=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\cdots+\frac{1}{n(n+1)(n+2)}, then 8S∞8S_\infty is
Sequences & Series · easy
If the sum to infinity of 1+4x+7x2+10x3+⋯1+4x+7x^2+10x^3+\cdots is 3516\frac{35}{16}, where ∣x∣<1|x|<1, then xx equals

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.

Sign up freeExplore doMath