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JEE Mathematics 2024 Question with Solution

The sum of the series:

(11−3⋅(12)+14)+(21−3⋅(22)+24)+(31−3⋅(32)+34)+…\left( \frac{1}{1 - 3\cdot(1^2) + 1^4} \right) + \left( \frac{2}{1 - 3\cdot(2^2) + 2^4} \right) + \left( \frac{3}{1 - 3\cdot(3^2) + 3^4} \right) + \ldots up to 1010 terms

  • A

    45109\frac{45}{109}

  • B

    −45109-\frac{45}{109}

  • C

    55109\frac{55}{109}

  • D

    −55109-\frac{55}{109}

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given:

Tr=r1−3r2+r4T_r = \frac{r}{1 - 3r^2 + r^4}

and we need the sum

S=∑r=110TrS = \sum_{r=1}^{10} T_r

Find: The sum of the first 1010 terms.

Rewrite the denominator:

1−3r2+r4=r4−3r2+1=r4−2r2+1−r2=(r2−1)2−r21 - 3r^2 + r^4 = r^4 - 3r^2 + 1 = r^4 - 2r^2 + 1 - r^2 = (r^2 - 1)^2 - r^2

So,

1−3r2+r4=(r2−r−1)(r2+r−1)1 - 3r^2 + r^4 = (r^2 - r - 1)(r^2 + r - 1)

Therefore,

Tr=r(r2−r−1)(r2+r−1)T_r = \frac{r}{(r^2 - r - 1)(r^2 + r - 1)}

Now use partial fraction decomposition:

Tr=12[1r2−r−1−1r2+r−1]T_r = \frac{1}{2}\left[\frac{1}{r^2 - r - 1} - \frac{1}{r^2 + r - 1}\right]

Since

r2+r−1=(r+1)2−(r+1)−1r^2 + r - 1 = (r+1)^2 - (r+1) - 1

we get a telescoping series:

∑r=110Tr=12∑r=110[1r2−r−1−1r2+r−1]\sum_{r=1}^{10} T_r = \frac{1}{2} \sum_{r=1}^{10} \left[\frac{1}{r^2 - r - 1} - \frac{1}{r^2 + r - 1}\right]

All intermediate terms cancel, leaving

S=12[112−1−1−1102+10−1]S = \frac{1}{2}\left[\frac{1}{1^2 - 1 - 1} - \frac{1}{10^2 + 10 - 1}\right]

Now evaluate:

12−1−1=−1,102+10−1=1091^2 - 1 - 1 = -1, \qquad 10^2 + 10 - 1 = 109

Hence,

S=12[−1−1109]=12[−110109]=−55109S = \frac{1}{2}\left[-1 - \frac{1}{109}\right] = \frac{1}{2}\left[-\frac{110}{109}\right] = -\frac{55}{109}

The solution working shown telescopes to −55109-\frac{55}{109}, so the matching option from the given list should be D. However, the extracted the solution also states Option D while one algebraic line incorrectly writes 55109\frac{55}{109} without the negative sign. The correct option is C? No—the option containing −55109-\frac{55}{109} is D.

Therefore, the correct option is D.

Telescoping Structure

Given:

Tr=r1−3r2+r4T_r = \frac{r}{1 - 3r^2 + r^4}

Find: Sum of the first 1010 terms.

Factor the denominator carefully:

1−3r2+r4=r4−3r2+11 - 3r^2 + r^4 = r^4 - 3r^2 + 1 =r4−2r2+1−r2= r^4 - 2r^2 + 1 - r^2 =(r2−1)2−r2= (r^2 - 1)^2 - r^2 =(r2−1−r)(r2−1+r)= (r^2 - 1 - r)(r^2 - 1 + r) =(r2−r−1)(r2+r−1)= (r^2 - r - 1)(r^2 + r - 1)

So,

Tr=r(r2−r−1)(r2+r−1)T_r = \frac{r}{(r^2 - r - 1)(r^2 + r - 1)}

Observe that the difference of the two denominator factors is

(r2+r−1)−(r2−r−1)=2r(r^2 + r - 1) - (r^2 - r - 1) = 2r

Hence,

1r2−r−1−1r2+r−1=2r(r2−r−1)(r2+r−1)\frac{1}{r^2 - r - 1} - \frac{1}{r^2 + r - 1} = \frac{2r}{(r^2 - r - 1)(r^2 + r - 1)}

Therefore,

Tr=12[1r2−r−1−1r2+r−1]T_r = \frac{1}{2}\left[\frac{1}{r^2 - r - 1} - \frac{1}{r^2 + r - 1}\right]

Now sum from r=1r=1 to 1010:

S=12∑r=110[1r2−r−1−1r2+r−1]S = \frac{1}{2} \sum_{r=1}^{10} \left[\frac{1}{r^2 - r - 1} - \frac{1}{r^2 + r - 1}\right]

Since

r2+r−1=(r+1)2−(r+1)−1r^2 + r - 1 = (r+1)^2 - (r+1) - 1

this becomes telescoping.

Write the first few terms:

S=12[(1−1−11)+(11−15)+(15−111)+⋯+(189−1109)]S = \frac{1}{2}\left[\left(\frac{1}{-1} - \frac{1}{1}\right) + \left(\frac{1}{1} - \frac{1}{5}\right) + \left(\frac{1}{5} - \frac{1}{11}\right) + \cdots + \left(\frac{1}{89} - \frac{1}{109}\right)\right]

All the middle terms cancel, leaving

S=12[1−1−1109]S = \frac{1}{2}\left[\frac{1}{-1} - \frac{1}{109}\right]

Thus,

S=12[−1−1109]=−55109S = \frac{1}{2}\left[-1 - \frac{1}{109}\right] = -\frac{55}{109}

Therefore, the sum is −55109-\frac{55}{109}, so the correct option is D.

Common mistakes

  • Factoring r4−3r2+1r^4 - 3r^2 + 1 incorrectly. A common error is to miss the identity (r2−1)2−r2(r^2 - 1)^2 - r^2. Rewrite it systematically before factorizing into (r2−r−1)(r2+r−1)(r^2 - r - 1)(r^2 + r - 1).

  • Dropping the factor 12\frac{1}{2} in partial fractions. The numerator difference is 2r2r, not rr. So the decomposition must be multiplied by 12\frac{1}{2}.

  • Missing the telescoping pattern by not recognizing that r2+r−1=(r+1)2−(r+1)−1r^2 + r - 1 = (r+1)^2 - (r+1) - 1. This shift is what causes cancellation across consecutive terms.

  • Sign error in the last step. Since the first surviving term is 1−1\frac{1}{-1}, the sum is negative. Recheck the final subtraction before choosing the option.

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