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Maths Question 5 – JEE-MAIN 2026

The sum 131+13+231+3+13+23+331+3+5+ up to 8 terms, is:

The problem involves a series where each term is a fraction. First, determine the general n-th term of the series.

Step 1: Determine the general term (Tn) of the series✦ Active

The numerator of the n-th term is the sum of the first n cubes, which is given by the formula k=1nk3=(n(n+1)2)2.

The denominator of the n-th term is the sum of the first n odd numbers, which is given by the formula k=1n(2k1)=n2.

Tn=(n(n+1)2)2n2=n2(n+1)24n2=(n+1)24
Step 2: Calculate the sum of the series up to 8 terms○ Expand

We need to find the sum S=n=18Tn. Substitute the expression for Tn:

S=n=18(n+1)24

Let k=n+1. When n=1, k=2. When n=8, k=9. The sum becomes:

S=14k=29k2
Step 3: Apply the sum of squares formula and find the final sum○ Expand

The sum of the first N squares is given by k=1Nk2=N(N+1)(2N+1)6. We can write k=29k2 as (k=19k2)12.

k=29k2=9(9+1)(29+1)612=9101961=35191=2851=284

Now, substitute this value back into the expression for S:

S=14284=71
💡 Teacher's Secret Hint

Ensure to subtract the k=1 term when the summation starts from k=2.

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