Let S n - k -1 n n - n 2- kn - k 2 and T n - k -0 n -1 n - n 2- kn - k 2 for a -1-2-3- - Then-A- Sn-3 -3- T n -3 -3

The correct options are A S_n lt; π/3√(3) D T_n gt; π/3√(3)We have ∑_k=1^nn/n^2 + kn + k^2 and T_n = ∑_k=0^n 1n/n^2 + kn + k^2 n = 1, 2, 3..... For n = ...

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Byju's Answer

Standard XI

Mathematics

Vn method

Let S n =∑ k ...

Question

Let $S_{n} = \sum_{k = 1}^{n} \frac{n}{n^{2} + k n + k^{2}} a n d T_{n} = \sum_{k = 0}^{n - 1} \frac{n}{n^{2} + k n + k^{2}}$ for a = 1, 2, 3,...... Then,

S_{n} < \frac{π}{3 \sqrt{3}}

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S_{n} > \frac{π}{3 \sqrt{3}}

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T_{n} < \frac{π}{3 \sqrt{3}}

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T_{n} > \frac{π}{3 \sqrt{3}}

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Solution

The correct options are
A $S_{n} < \frac{π}{3 \sqrt{3}}$
D $T_{n} > \frac{π}{3 \sqrt{3}}$
We have $\sum_{k = 1}^{n} \frac{n}{n^{2} + k n + k^{2}} a n d T_{n} = \sum_{k = 0}^{n - 1} \frac{n}{n^{2} + k n + k^{2}} n = 1, 2, 3..... F o r n = 1 w e g e t S_{1} = \frac{1}{1 + 1 + 1} = \frac{1}{3} = 0.3 a n d T_{1} = \frac{1}{1 + 0} = 1 A l s o \frac{π}{3 \sqrt{3}} = \frac{π \sqrt{3}}{9} = \frac{3.14 \times 1.73}{9} = 0.34 \times 1.73 = 0.58 ∴ S_{1} < \frac{π}{3 \sqrt{3}} < T_{1}, ∴ S_{n} < \frac{π}{3 \sqrt{3}} a n d T_{n} > \frac{π}{3 \sqrt{3}}$

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Similar questions

Q. Let

S_{n} = \sum_{k = 1}^{n} \frac{n}{n^{2} + k n + k^{2}} a n d T_{n} = \sum_{k = 0}^{n - 1} \frac{n}{n^{2} + k n + k^{2}}

for a = 1, 2, 3,...... Then,

Q. Let

S_{n} = \sum_{k = 1}^{n} \frac{n}{n^{2} + k n + k^{2}}, T_{n} = n - 1 \sum k = 0 \frac{n}{n^{2} + k n + k^{2}}

for

n = 1, 2, 3...

Then

Q. Let

S_{n} = n \sum k = 1 \frac{n}{n^{2} + k n + k^{2}}

and

T_{n} = n - 1 \sum k = 0 \frac{n}{n^{2} + k n + k^{2}}

for

n = 1, 2, 3, \dots

. Then,

Q. Let

S_{n} = n \sum k = 1 \frac{n}{n^{2} + k n + k^{2}}

and

T_{n} = n - 1 \sum k = 0 \frac{n}{n^{2} + k n + k^{2}}

for

n = 1, 2, 3, . .

Then,

Q. Let

S_{n} = n \sum k = 1 \frac{k}{(k - 1)^{4 / 3} + (k^{2} - 1)^{2 / 3} + (k + 1)^{4 / 3}}

and

lim n \to \infty \frac{S_{n}}{n^{2 / 3}} = \frac{1}{p} .

Then the value of

p

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Let Sn=∑nk=1nn2+kn+k2 and Tn=∑n−1k=0nn2+kn+k2 for a = 1, 2, 3,...... Then,

The correct options are A Sn<π3√3 D Tn>π3√3We have ∑nk=1nn2+kn+k2and Tn=∑n−1k=0nn2+kn+k2n=1,2,3.....For n=1 we getS1=11+1+1=13=0.3 and T1=11+0=1Also π3√3=π√39=3.14×1.739=0.34×1.73=0.58∴S1<π3√3<T1,∴Sn<π3√3 and Tn>π3√3

Let $S_{n} = \sum_{k = 1}^{n} \frac{n}{n^{2} + k n + k^{2}} a n d T_{n} = \sum_{k = 0}^{n - 1} \frac{n}{n^{2} + k n + k^{2}}$ for a = 1, 2, 3,...... Then,