Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Sequences and Series question

2022 · 28 Jun · Shift 2 · Q45
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Mathematics
  4. /Sequences and Series
  5. /2022 · 28 Jun · Shift 2 · Q45

Sequences and Series question

2022 · 28 Jun · Shift 2 · Q45

JEE MainMathematicsSequences and SeriesNumerical+4 / −1
Let for n = 1, 2, ......, 50, Sn be the sum of the infinite geometric progression whose first term is n2 and whose common ratio is 1(n+1)2{1 \over {{{(n + 1)}^2}}}(n+1)21​. Then the value of 126+∑n=150(Sn+2n+1−n−1){1 \over {26}} + \sum\limits_{n = 1}^{50} {\left( {{S_n} + {2 \over {n + 1}} - n - 1} \right)}261​+n=1∑50​(Sn​+n+12​−n−1) is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 41651

  1. Write the sum of the infinite G.P.

For each n=1,2,…,50n=1,2,\dots,50n=1,2,…,50:

  • First term a=n2a=n^2a=n2
  • Common ratio r=1(n+1)2r=\dfrac{1}{(n+1)^2}r=(n+1)21​

Since ∣r∣<1|r|<1∣r∣<1, the sum to infinity is

Sn=a1−r=n21−1(n+1)2.S_n=\frac{a}{1-r}=\frac{n^2}{1-\frac{1}{(n+1)^2}}.Sn​=1−ra​=1−(n+1)21​n2​.

Now simplify:

1−1(n+1)2=(n+1)2−1(n+1)2=n2+2n(n+1)2=n(n+2)(n+1)2.1-\frac{1}{(n+1)^2}=\frac{(n+1)^2-1}{(n+1)^2}=\frac{n^2+2n}{(n+1)^2}=\frac{n(n+2)}{(n+1)^2}.1−(n+1)21​=(n+1)2(n+1)2−1​=(n+1)2n2+2n​=(n+1)2n(n+2)​.

Hence,

Sn=n2⋅(n+1)2n(n+2)=n(n+1)2n+2.S_n=n^2\cdot \frac{(n+1)^2}{n(n+2)}=\frac{n(n+1)^2}{n+2}.Sn​=n2⋅n(n+2)(n+1)2​=n+2n(n+1)2​.
  1. Simplify the expression inside the summation

We need

Sn+2n+1−n−1.S_n+\frac{2}{n+1}-n-1.Sn​+n+12​−n−1.

Substitute SnS_nSn​:

n(n+1)2n+2−(n+1)+2n+1.\frac{n(n+1)^2}{n+2}-(n+1)+\frac{2}{n+1}.n+2n(n+1)2​−(n+1)+n+12​.

First simplify

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

So

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

A cleaner way is to divide SnS_nSn​ first:

Sn=n(n+1)2n+2=n2−1+4n+2−4?S_n=\frac{n(n+1)^2}{n+2}=n^2-1+\frac{4}{n+2}-\frac{4}{?}Sn​=n+2n(n+1)2​=n2−1+n+24​−?4​

Instead, do polynomial division directly on the whole main part:

\frac{n(n+1)^2}{n+2}= rac{n^3+2n^2+n}{n+2}.

Divide:

n3+2n2+nn+2=n2+nn+2=n2+1−2n+2.\frac{n^3+2n^2+n}{n+2}=n^2+\frac{n}{n+2}=n^2+1-\frac{2}{n+2}.n+2n3+2n2+n​=n2+n+2n​=n2+1−n+22​.

Thus

Sn=n2+1−2n+2.S_n=n^2+1-\frac{2}{n+2}.Sn​=n2+1−n+22​.

Therefore,

Sn+2n+1−n−1=(n2+1−2n+2)+2n+1−n−1=n2−n+2n+1−2n+2.S_n+\frac{2}{n+1}-n-1 =\left(n^2+1-\frac{2}{n+2}\right)+\frac{2}{n+1}-n-1 =n^2-n+\frac{2}{n+1}-\frac{2}{n+2}.Sn​+n+12​−n−1=(n2+1−n+22​)+n+12​−n−1=n2−n+n+12​−n+22​.

So the required expression is

126+∑n=150(n2−n+2n+1−2n+2).\frac{1}{26}+\sum_{n=1}^{50}\left(n^2-n+\frac{2}{n+1}-\frac{2}{n+2}\right).261​+n=1∑50​(n2−n+n+12​−n+22​).
  1. Split the summation
∑n=150(n2−n)+∑n=150(2n+1−2n+2)+126.\sum_{n=1}^{50}(n^2-n)+\sum_{n=1}^{50}\left(\frac{2}{n+1}-\frac{2}{n+2}\right)+\frac{1}{26}.n=1∑50​(n2−n)+n=1∑50​(n+12​−n+22​)+261​.
  1. Evaluate ∑(n2−n)\sum (n^2-n)∑(n2−n)
∑n=150(n2−n)=∑n=150n2−∑n=150n.\sum_{n=1}^{50}(n^2-n)=\sum_{n=1}^{50}n^2-\sum_{n=1}^{50}n.n=1∑50​(n2−n)=n=1∑50​n2−n=1∑50​n.

Using formulas:

∑n=150n=50⋅512=1275,\sum_{n=1}^{50} n = \frac{50\cdot 51}{2}=1275,n=1∑50​n=250⋅51​=1275, ∑n=150n2=50⋅51⋅1016=42925.\sum_{n=1}^{50} n^2 = \frac{50\cdot 51\cdot 101}{6}=42925.n=1∑50​n2=650⋅51⋅101​=42925.

Hence,

∑n=150(n2−n)=42925−1275=41650.\sum_{n=1}^{50}(n^2-n)=42925-1275=41650.n=1∑50​(n2−n)=42925−1275=41650.
  1. Evaluate the telescoping sum
∑n=150(2n+1−2n+2)=2∑n=150(1n+1−1n+2).\sum_{n=1}^{50}\left(\frac{2}{n+1}-\frac{2}{n+2}\right) =2\sum_{n=1}^{50}\left(\frac{1}{n+1}-\frac{1}{n+2}\right).n=1∑50​(n+12​−n+22​)=2n=1∑50​(n+11​−n+21​).

This telescopes:

=2(12−152)=1−126=2526.=2\left(\frac12-\frac{1}{52}\right)=1-\frac{1}{26}=\frac{25}{26}.=2(21​−521​)=1−261​=2625​.
  1. Add the extra term 126\frac{1}{26}261​

Total value is

41650+2526+126=41650+1=41651.41650+\frac{25}{26}+\frac{1}{26}=41650+1=41651.41650+2625​+261​=41650+1=41651.
  1. Final answer
41651\boxed{41651}41651​
PreviousNext

More from Sequences and Series

  • Let a1​,a2​,a3​,… be an A.P. If r=1∑∞​2rar​​=4, then 4a2​ is equal to ​.2022 · Numerical
  • Let 3, 6, 9, 12, ....... upto 78 terms and 5, 9, 13, 17, ...... upto 59 terms be two series. Then, the sum of the terms common to both the series is equal to ​.2022 · Numerical
  • Let a1, a2, ..........., a21 be an AP such that n=1∑20​an​an+1​1​=94​. If the sum of this AP is 189, then a6a16 is equal to :2021 · MCQ
  • Consider an arithmetic series and a geometric series having four initial terms from the set {11, 8, 21, 16, 26, 32, 4}. If the last terms of these series are the maximum possible four digit numbers, then the number of common terms in these…2021 · Numerical
  • Let 161​, a and b be in G.P. and a1​, b1​, 6 be in A.P., where a, b > 0. Then 72(a + b) is equal to ​.2021 · Numerical
  • Let S1 be the sum of first 2n terms of an arithmetic progression. Let S2 be the sum of first 4n terms of the same arithmetic progression. If (S2 − S1) is 1000, then the sum of the first 6n terms of the arithmetic progression is equal to :2021 · MCQ
  • Let Sn denote the sum of first n-terms of an arithmetic progression. If S10 = 530, S5 = 140, then S20 − S6 is equal to:2021 · MCQ
  • The sum of all the elements in the set {n ∈ {1, 2, ....., 100} | H.C.F. of n and 2040 is 1} is equal to ​.2021 · Numerical