Summation MCQ Quiz in मल्याळम - Objective Question with Answer for Summation - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept:

In the power series un = an xn;

$\underset{n\to \mathrm{\infty }}{lim}\left|\frac{u_{n+1}}{u_n}\right|=\underset{n\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}\left|\left(\frac{a_{n+1}}{a_n}\right)\right|.x$

If $\underset{n\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}\left(\frac{a_{n+1}}{a_n}\right)=l,$ then by ratio test, the series converges, when |lx| is numerically less than 1 i.e., when |x| < 1/l and diverges for other values.

The interval (– 1/l) < x < (1/l) is called the interval of convergence of the power series.

We have to check the convergence at the boundary points also.

Calculation:

Given series is $\sum u_n=\frac{x^{2n}}{\left(n+1\right)\sqrt{n}}$ 

We have $u_n=\frac{x^{2n}}{\left(n+1\right)\sqrt{n}}$ and $u_{n+1}=\frac{x^{2n+2}}{\left(n+2\right)\sqrt{n+1}}$ 

$\Rightarrow \frac{u_{n+1}}{u_n}=x^2\frac{n+1}{n+2}\sqrt{\frac{n}{n+1}}$

Now from the ratio test,

$\underset{n\to \mathrm{\infty }}{lim}\left|\frac{u_{n+1}}{u_n}\right|=\underset{n\to \mathrm{\infty }}{lim}\left|x^2\frac{n+1}{n+2}\sqrt{\frac{n}{n+1}}\right|$

$\Rightarrow x^2\underset{n\to \mathrm{\infty }}{lim}\left|\frac{1+\frac{1}{n}}{1+\frac{2}{n}}\sqrt{\frac{1}{1+\frac{1}{n}}}\right|$

⇒ x²;

Therefore the series converges for x² < 1 ⇒ -1 < x < 1

Ratio test fails when x² = 1, therefore at x² = 1;

At x² = 1, $u_n=\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{1}{n^{\frac{3}{2}}}.\frac{1}{1+\frac{1}{n}}$ 

Let’s take $v_n=\frac{1}{n^{3/2}}$ 

We get $\underset{n\to \mathrm{\infty }}{lim}\frac{u_n}{v_n}=\underset{n\to \mathrm{\infty }}{lim}\frac{1}{1+\frac{1}{n}}=1,afinitequantity$ 

Therefore, both series will converge or diverge together.

We already know v_n is convergent (p-series) ⇒ u_n will also be convergent;

⇒ at x² = 1, the series ∑u_n is convergent;

Concept:

The Ratio test:

Theorem: Let ∑a_n be a series and If $\underset{n\to \mathrm{\infty }}{lim}\frac{a_{k+1}}{a_k}=L$, and

if L < 1 then the series converges, but if L > 1 the series diverges.

If L = 1, then the test is inconclusive, then we have to go for another test.

Concept used: 

Suppose  f(x) is positive decreasing continuous function on the interval [1, ∞ ) with f(n) = a_n 

then a_n  is converging iff $\underset{1}{\overset{\infty }{\int }}$f(x) dx converges.

Calculations:

$\underset{1}{\overset{\infty }{\int }}\varphi (x)dx=\underset{1}{\overset{\infty }{\int }}\frac{2x^3}{3x^4+2}dx$

$=\frac{1}{6}\underset{1}{\overset{\infty }{\int }}\frac{dt}{t}=\frac{1}{6}{\left[\log t\right]}_5^{\infty }=\infty$

[Where t = 3x⁴ + 2]

By integral test, ∑u_n is divergent.

Concept:

A series in which the terms are alternatively positive or negative is called an alternating series.

Liebnitz’s series:

An alternating series u₁ – u₂ + u₃ – u₄ + … converges if

(i) Each term is numerically less than its preceding term,

(ii) $\underset{n\to \mathrm{\infty }}{lim}{u}_n=0$

If $\underset{n\to \mathrm{\infty }}{lim}{u}_n\ne 0$, the given series is oscillatory.

Calculation:

Given series is $\sum _{n=1}^{\mathrm{\infty }}\frac{{\left(-1\right)}^{n-1}n}{2n-1}$

Now $u_n=\frac{n}{2n-1}$

⇒ $u_n-u_{n-1}=\frac{n}{2n-1}-\frac{n-1}{2n-3}=\frac{-1}{\left(2n-1\right)\left(n-3\right)}$

So each term is numerically less than its preceding term.

Now limit,

$\underset{n\to \mathrm{\infty }}{lim}{u}_n=\underset{n\to \mathrm{\infty }}{lim}\frac{n}{2n-1}=\frac{1}{2}$

⇒ $\underset{n\to \mathrm{\infty }}{lim}{u}_n\ne 0$

∴ Series is oscillatory

Given:
Series 1 = $1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...$

Series 2 = $2+2\times \frac{1}{3}+2\times \frac{1}{3^2}+2\times \frac{1}{3^3}+...$

...

Series n = $n+n\times \frac{1}{n+1}+n\times \frac{1}{(n+1{)}^2}+n\times \frac{1}{(n+1{)}^3}+...$

Concept:
Sum of an infinite G.P. series, S_∞ = $\frac{a}{1-r}$ , when |r| < 1

Sum of an A.P. series, S_AP = $\frac{n(a+l)}{2}$ , where, $l$ = last term of series

Calculation:

For Series 1, a = 1, and r = $\frac{1}{2}$
∴ S₁ = $\frac{a}{1-r}$ = $\frac{1}{1-\frac{1}{2}}$
⇒ S₁ = 2

Similarly, for Series 2, a = 2 and r = $\frac{1}{3}$
∴ S₁ = $\frac{2}{1-\frac{1}{3}}$
⇒ S₂ = 3

Similarly,

S3 = 4

S₄ = 5, ...

S_n = n + 1

So, S₁, S₂, S₃, ... , Sn is an Arithmetic Progression (A.P.),

For A.P.,

a = S1 = 2

n = n
$l$ = Sn = n + 1,

Therefore,
(S1 + S2 + S3 + ... + S_n) = $\frac{n(a+l)}{2}=\frac{n(2+n+1)}{2}$

∴ ( S1 + S2 + S3 + ... + S_n ) = $\frac{n(n+3)}{2}$

GIVEN:

First term a = – 9

Last term l = 51

Number of term n = 16

FORMULAE USED:

Sum = n/2 × (a + l)

CALCULATION:

⇒ 16/2 × (- 9 + 51) = sum

Concept:

In the power series un = an xn;

$\underset{n\to \mathrm{\infty }}{lim}\left|\frac{u_{n+1}}{u_n}\right|=\underset{n\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}\left|\left(\frac{a_{n+1}}{a_n}\right)\right|.x$

If $\underset{n\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}\left(\frac{a_{n+1}}{a_n}\right)=l,$ then by ratio test, the series converges, when |lx| is numerically less than 1 i.e., when |x| < 1/l and diverges for other values.

The interval (– 1/l) < x < (1/l) is called the interval of convergence of the power series.

We have to check the convergence at the boundary points also.

Useful concept:

For positive terms series, if limit (n → ∞) u_n = 0, then it is convergent, if not equal to zero, it is divergent.

Calculation: 

Given series is $1+\frac{2}{5}x+\frac{6}{9}{x}^2+\dots +\frac{2^n-2}{2^n+1}{x}^{n-1}+\dots (x>0)$

As x > 0, it is a positive term series, so ratio test can be employed.

Now we have

$u_n=\frac{2^n-2}{2^n+1}{x}^{n-1},u_{n+1}=\frac{2^{n+1}-2}{2^{n+1}+1}{x}^n$

$\Rightarrow \frac{u_n}{u_{n+1}}=\frac{2^n-2}{2^n+1}.\frac{2^{n+1}+1}{2^{n+1}-2}.\frac{x^{n-1}}{x^n}$

$\Rightarrow \frac{u_n}{u_{n+1}}=\frac{1-\frac{2}{2^n}}{1+\frac{1}{2^n}}.\frac{2+\frac{1}{2^n}}{2-\frac{1}{2^n}}.\frac{1}{x}$

Now

$\underset{n\to \mathrm{\infty }}{lim}\frac{u_n}{u_{n+1}}=\frac{1-0}{1+0}.\frac{2+0}{2-0}.\frac{1}{x}=\frac{1}{x}$

By ratio test, series converges if limit is greater than 1 and diverges if less than 1.

Now the given series converges for 1/x > 1 ⇒ x < 1;

Series diverges for 1/x < 1 ⇒ x > 1;

At x = 1, the test fails,

When x = 1, $\underset{n\to \mathrm{\infty }}{lim}{u}_n=\underset{n\to \mathrm{\infty }}{lim}\frac{2^n-2}{2^n+1}=\underset{n\to \mathrm{\infty }}{lim}\frac{1-\frac{2}{2^n}}{1+\frac{1}{2^n}}=1\ne 0$

⇒ ∑u_n diverges for x = 1

Hence the given series converges for x < 1 and diverges for x ≥ 1;

Concept:

Cauchy’s Root Test:

In a positive series ∑u_n, if $\underset{n\to \mathrm{\infty }}{lim}{\left(u_n\right)}^{\frac{1}{n}}=\lambda$

Then the series converges for λ < 1, and diverges for λ > 1; 

Calculation:

Given series is ${\left(\frac{2^2}{1^2}-\frac{2}{1}\right)}^{-1}+{\left(\frac{3^3}{2^3}-\frac{3}{2}\right)}^{-2}+{\left(\frac{4^4}{3^3}-\frac{4}{3}\right)}^{-3}+\dots \mathrm{\infty }$

The general form will be 

∑u_n ⇒ $\sum {\left[\frac{{\left(n+1\right)}^{n+1}}{n^{n+1}}-\frac{n+1}{n}\right]}^{-n}$

⇒ $u_n={\left[\frac{{\left(n+1\right)}^{n+1}}{n^{n+1}}-\frac{n+1}{n}\right]}^{-n}$

Now

$u_n^{\frac{1}{n}}={\left[\frac{{\left(n+1\right)}^{n+1}}{n^{n+1}}-\frac{n+1}{n}\right]}^{-1}={\left(1+\frac{1}{n}\right)}^{-1}{\left[{\left(1+\frac{1}{n}\right)}^n-1\right]}^{-1}$

$\underset{n\to \mathrm{\infty }}{lim}{\left(u_n\right)}^{\frac{1}{n}}=\underset{n\to \mathrm{\infty }}{lim}1.{\left[{\left(1+\frac{1}{n}\right)}^n-1\right]}^{-1}$

Using the standard limit $\underset{n\to \mathrm{\infty }}{lim}{\left(1+\frac{1}{n}\right)}^n=e$

⇒ $\underset{n\to \mathrm{\infty }}{lim}{\left(u_n\right)}^{\frac{1}{n}}=\frac{1}{e-1}<1$

Since the limit is less than one, the given series is convergent.

The given alternating series is:

$x-\frac{x^2}{2^2}+\frac{x^3}{3^2}-\frac{x^4}{4^2}+\dots \mathrm{\infty }$

$u_n=\frac{{\le ft(-1}^{n-1}{x}^n}{n^2}$

$u_{n+1}=\frac{{\le ft(-1)}^n{x}^{n+1}}{{\le ft(n+1)}^2}$

$\le ft|u_n|=\frac{{\le ft|x|}^n}{n^2}$

$\le ft|u_{n+1}|=\frac{{\le ft|x}^{n+1}}{{\le ft(n+1)}^2}$

$\frac{\le ft|u_n|}{\le ft|u_{n+1}|}=\frac{\frac{{\le ft|x|}^n}{n^2}}{\frac{{\le ft|x|}^{n+1}}{{\le ft(n+1)}^2}}=\frac{{\le ft(n+1)}^2}{n^2}\times \frac{1}{\le ft|x|}$

Now, $\underset{n\to \mathrm{\infty }}{lim}\frac{\le ft|u_n|}{\le ft|u_{n+1}|}=\underset{n\to \mathrm{\infty }}{lim}\frac{{\le ft(n+1)}^2}{n^2}\times \frac{1}{\le ft|x|}=\frac{1}{\le ft|x|}$

Thus the corresponding absolute series is convergent for $\frac{1}{\le ft|x|}>1$ i.e. for |x| < 1 i.e. -1 < x < 1

Absolute series is divergent for $\frac{1}{\le ft|x|}<1$ i.e. for |x| > 1 i.e. x > 1 and x < -1

Since absolute convergence implies convergence, the given infinite series is convergent for -1 < x < 1

For x = 1, the absolute series becomes

$\sum _{n=1}^{\mathrm{\infty }}\le ft|u_n|=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2}$

$\le ft|u_n|=\frac{1}{n^2}$

$\le ft|u_{n+1}|=\frac{1}{{\le ft(n+1)}^2}$

n < (n+1), for all n

$\Rightarrow n^2<\le ft(n+1{)}^2$

$\Rightarrow \frac{1}{{\le ft(n+1)}^2}<\frac{1}{n^2}$

$\le ft|u_{n+1}|<\le ft|u_n|$; for all n

Moreover, $\underset{n\to \mathrm{\infty }}{lim}\le ft|u_n|=\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n^2}=0$

Thus both the conditions of Leibnitz rule re satisfied

So the given series is convergent for x = 1

For x = -1, the absolute series becomes

$\sum _{n=1}^{\mathrm{\infty }}{u}_n=\sum _{n=1}^{\mathrm{\infty }}\frac{-1}{n^2}$

It is evident that the series $\sum _{n=1}^{\mathrm{\infty }}\frac{-1}{n^2}$ is convergent as it is a p series with power greater than 1.

Concept Used:-

If the summation from 0 to n such that ${\displaystyle \sum _{\mathrm{r}=0}^{\mathrm{n}}}$ ar xr is given, then the expanded form of it can be written as,

 ⇒ ${\displaystyle \sum _{\mathrm{r}=0}^{\mathrm{n}}}$ ar xr = a₀+a₁ x+a₂ x²+a₃ x³+a₄ x⁴+..........+a_nxⁿ

Explanation:-

Given,

(1+ x + x2)n = ${\displaystyle \sum _{\mathrm{r}=0}^{2\mathrm{n}}}$ ar xr, 

On the expanding right-hand side, we get,

${(1+x+x^2)}^n=a_0+a_{1}x+a_2{x}^2+a_3{x}^3+a_4{x}^4+\cdots +a_{2n}{x}^{2n}$

Differentiate it with respect to x,

$n{(1+x+x^2)}^{n-1}(1+2x)=a_1+2a_{2}x+3a_3{x}^2+4a_4{x}^3+\cdots +2n{a}_n{x}^{2x-1}$Now put x = -1 in the above equation,

$$\begin{gathered} \Rightarrow n(1-1+1{)}^{n-1}(1-2)=a_1-2a_2+3a_3-4a_4+5a_5\cdots -2n{a}_{2n} \\ \Rightarrow n^1(1{)}^{n-1}(-1)=a_1-2a_2+3a_3-4a_4+5a_5\cdots -2n{a}_{2n} \\ \Rightarrow -n=a_1-2a_2+3a_3-4a_4+5a_5\cdots -2n{a}_{2n} \end{gathered}$$

So, the value of a1 − 2a2 + 3a3 − …. −2n a2n is -n.

Hence, the correct option is 3.