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Sequences and Series question

2024 · 31 Jan · Shift 1 · Q40
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  5. /2024 · 31 Jan · Shift 1 · Q40

Sequences and Series question

2024 · 31 Jan · Shift 1 · Q40

JEE MainMathematicsSequences and SeriesMCQ+4 / −1
The sum of the series 11−3⋅12+14+21−3⋅22+24+31−3⋅32+34+…\frac{1}{1-3 \cdot 1^2+1^4}+\frac{2}{1-3 \cdot 2^2+2^4}+\frac{3}{1-3 \cdot 3^2+3^4}+\ldots1−3⋅12+141​+1−3⋅22+242​+1−3⋅32+343​+… up to 10 -terms is
  1. A
    45109\frac{45}{109}10945​
  2. B
    −55109-\frac{55}{109}−10955​
  3. C
    55109\frac{55}{109}10955​
  4. D
    −45109-\frac{45}{109}−10945​
View written solutionFree

Correct answer: B

  1. Write the general term

The given series is

∑n=110n1−3n2+n4.\sum_{n=1}^{10} \frac{n}{1-3n^2+n^4}.n=1∑10​1−3n2+n4n​.

So the general term is

Tn=nn4−3n2+1.T_n = \frac{n}{n^4-3n^2+1}.Tn​=n4−3n2+1n​.
  1. Factor the denominator

Observe that

n4−3n2+1=(n2−n−1)(n2+n−1).n^4-3n^2+1=(n^2-n-1)(n^2+n-1).n4−3n2+1=(n2−n−1)(n2+n−1).

Because

(n2−n−1)(n2+n−1)=((n2−1)−n)((n2−1)+n)=(n2−1)2−n2=n4−3n2+1.(n^2-n-1)(n^2+n-1)=((n^2-1)-n)((n^2-1)+n)=(n^2-1)^2-n^2=n^4-3n^2+1.(n2−n−1)(n2+n−1)=((n2−1)−n)((n2−1)+n)=(n2−1)2−n2=n4−3n2+1.

Hence

Tn=n(n2−n−1)(n2+n−1).T_n=\frac{n}{(n^2-n-1)(n^2+n-1)}.Tn​=(n2−n−1)(n2+n−1)n​.
  1. Split into partial fractions in telescoping form

Consider

12(1n2−n−1−1n2+n−1).\frac{1}{2}\left(\frac{1}{n^2-n-1}-\frac{1}{n^2+n-1}\right).21​(n2−n−11​−n2+n−11​).

Its numerator becomes

12⋅(n2+n−1)−(n2−n−1)(n2−n−1)(n2+n−1)=12⋅2n(n2−n−1)(n2+n−1)=n(n2−n−1)(n2+n−1).\frac{1}{2}\cdot \frac{(n^2+n-1)-(n^2-n-1)}{(n^2-n-1)(n^2+n-1)} =\frac{1}{2}\cdot \frac{2n}{(n^2-n-1)(n^2+n-1)} =\frac{n}{(n^2-n-1)(n^2+n-1)}.21​⋅(n2−n−1)(n2+n−1)(n2+n−1)−(n2−n−1)​=21​⋅(n2−n−1)(n2+n−1)2n​=(n2−n−1)(n2+n−1)n​.

Thus

Tn=12(1n2−n−1−1n2+n−1).T_n=\frac{1}{2}\left(\frac{1}{n^2-n-1}-\frac{1}{n^2+n-1}\right).Tn​=21​(n2−n−11​−n2+n−11​).

Now note that

n2+n−1=(n+1)2−(n+1)−1.n^2+n-1=(n+1)^2-(n+1)-1.n2+n−1=(n+1)2−(n+1)−1.

So if we define

an=1n2−n−1,a_n=\frac{1}{n^2-n-1},an​=n2−n−11​,

then

1n2+n−1=an+1.\frac{1}{n^2+n-1}=a_{n+1}.n2+n−11​=an+1​.

Therefore,

Tn=12(an−an+1).T_n=\frac{1}{2}(a_n-a_{n+1}).Tn​=21​(an​−an+1​).
  1. Apply telescoping sum

Hence

S10=∑n=110Tn=12∑n=110(an−an+1).S_{10}=\sum_{n=1}^{10} T_n =\frac{1}{2}\sum_{n=1}^{10}(a_n-a_{n+1}).S10​=n=1∑10​Tn​=21​n=1∑10​(an​−an+1​).

This telescopes to

S10=12(a1−a11).S_{10}=\frac{1}{2}(a_1-a_{11}).S10​=21​(a1​−a11​).

Now,

a1=112−1−1=1−1=−1,a_1=\frac{1}{1^2-1-1}=\frac{1}{-1}=-1,a1​=12−1−11​=−11​=−1,

and

a11=1112−11−1=1121−11−1=1109.a_{11}=\frac{1}{11^2-11-1}=\frac{1}{121-11-1}=\frac{1}{109}.a11​=112−11−11​=121−11−11​=1091​.

So,

S10=12(−1−1109)=12(−110109)=−55109.S_{10}=\frac{1}{2}\left(-1-\frac{1}{109}\right) =\frac{1}{2}\left(-\frac{110}{109}\right) =-\frac{55}{109}.S10​=21​(−1−1091​)=21​(−109110​)=−10955​.
  1. Check options

The value is

−55109,-\frac{55}{109},−10955​,

which matches Option B.

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