Are Sequence and Series solutions available in PDF format?

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9 months ago

Yes, NCERT Solutions for Class 11 Maths Chapter 8 Sequences and Series are readily available in PDF format. These NCERT Solution PDF for Sequence and Series are designed for students to get enough practice problems to solve and offline access of the resource to practice whenevre they want.

Shiksha p

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Similar Questions for you

If U_n = $(1 + \frac{1}{n^{2}}) {(1 + \frac{2^{2}}{n^{2}})}^{2} . . . . . . {(1 + \frac{n^{2}}{n^{2}})}^{n}, t h e n \underset{n \to \infty}{l i m} {(U_{n})}^{\frac{- 4}{n^{2}}}$  is equal to

Vishal Baghel

$l = \underset{n \to \infty}{l i m} {(u_{n})}^{\frac{- 4}{n^{2}}} = \underset{n \to \infty}{l i m} {(1 + \frac{1}{n^{2}}) {(1 + \frac{2^{2}}{n^{2}})}^{2} {(1 + \frac{3^{2}}{n^{2}})}^{3} . . . . . . {(1 + \frac{n^{2}}{n^{2}})}^{n}}^{\frac{- 4}{n^{2}}},$ Taking log on both the sides

$\underset{n \to \infty}{l i m} \frac{- 4}{n^{2}} \sum_{r = 1}^{n} r l o g (1 + \frac{r^{2}}{n^{2}}) = \underset{n \to \infty}{l i m} \frac{- 4}{n} \sum_{r = 1}^{n} \frac{r}{n} l o g (1 + {(\frac{r}{n})}^{2})$

$∴ l o g l = - 4 \int_{0}^{1} x l o g (1 + x^{2}) d x$

$l o g l = - 2 [l o g 4 - 1] = - 2 l o g \frac{4}{e} = l o g \frac{e^{2}}{16}$

$l = \frac{e^{2}}{16}$

Vishal Baghel

$S_{n} = {z \in C : | z - 3 + 2 i | = \frac{n}{4}}$

represents a circle with centre C1 (3, 2) and radius

$r_{1} = \frac{n}{4}$

Similarly Tn represents circle with centre C2 (2, 3) and radius

$r_{1} = \frac{1}{n}$

$\sqrt[]{2} < | \frac{n}{4} - \frac{1}{n} |$

n take infinite values.

Vishal Baghel

$α_{n} = 1 9^{n} - 1 2^{n}$

$\frac{31 α_{9} - α_{10}}{57 . α_{2}} = \frac{31 (1 9^{9} - 1 2^{9}) - (1 9^{10} - 1 2^{10})}{57 (1 9^{8} - 1 2^{8})}$

$= \frac{1 9^{9} (31 - 19) - 1 2^{9} (31 - 12)}{57 (1 9^{8} - 1 2^{8})}$

$= \frac{1 9^{9} . 12 - 1 2^{9} . 19}{57 (1 9^{8} - 1 2^{8})} = 4$

Let ${a_{n}}_{n = 0}^{\infty}$ be a sequence such that a0 = a1 = 0 and an+2 = 2an+1 – an + 1 for all $n \geq 0 .$ Then, $\sum_{n = 2}^{\infty} \frac{a_{n}}{7^{n}}$ is equal to:

Alok Kumar Singh

$a_{n + 2} = 2 \cdot a_{n + 1} - a_{n + 1}$

$a_{2} = 2 a_{1} - a_{0} + 1$

a₂ = 1

a₃ = 3

a₄ = 6

So for n $\geq 2, a_{n} = \frac{n (n - 1)}{2}$

$\sum_{n = 2}^{} \frac{n (n - 1)}{2 \cdot 7^{n}} = \frac{1}{7^{2}} + \frac{3}{7^{3}} + \frac{6}{7^{4}} + . . . . . .$

Let S = $\frac{1}{7^{2}} + \frac{3}{7^{3}} + \frac{6}{7^{4}} + . . . . .$

$\frac{\frac{S}{7} = \frac{1}{7^{3}} + \frac{3}{7^{4}} + . . . .}{S - \frac{S}{7} = \frac{1}{7^{2}} + \frac{2}{7^{3}} + \frac{3}{7^{4}} + . . . .}$

$\frac{\frac{6}{7} S \cdot \frac{1}{7} = \frac{1}{7^{3}} + \frac{2}{7^{3}} + . . . .}{\frac{6}{7} S - \frac{6}{49} S = \frac{1}{7^{2}} + \frac{1}{7^{3}} + . . . .}$