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

Concept:

In the power series un = an xn;

\(\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{u_{n + 1}}}}{{{u_n}}}} \right| = \mathop {{\rm{lim}}}\limits_{n \to \infty } \left| {\left( {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right)} \right|.x\)

If \(\mathop {{\rm{lim}}}\limits_{n \to \infty } \left( {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right) = l,\) 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,

\(\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{u_{n + 1}}}}{{{u_n}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {{x^2}\frac{{n + 1}}{{n + 2}}\sqrt {\frac{n}{{n + 1}}} } \right|\)

\( \Rightarrow {x^2}\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{1 + \frac{1}{n}}}{{1 + \frac{2}{n}}}\sqrt {\frac{1}{{1 + \frac{1}{n}}}} } \right|\)

⇒ 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 \(\mathop {\lim }\limits_{n \to \infty } \frac{{{u_n}}}{{{v_n}}} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{{1 + \frac{1}{n}}} = 1,\;a\;finite\;quantity\) 

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 \(\mathop {\lim }\limits_{n \to \infty } \frac{{{a_{k + 1}}}}{{{a_k}}} = L\), 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 \(\int\limits_1^∞\)f(x) dx converges.

Calculations:

\(\int\limits_1^∞ {\phi (x)dx = } \int\limits_1^∞ {\frac{{2{x^3}}}{{3{x^4} + 2}}} dx\)

\( = \frac{1}{6}\int\limits_1^∞ {\frac{{dt}}{t}} = \frac{1}{6}\left[ {\log t} \right]_5^∞ = ∞ \)

[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) \(\mathop {\lim }\limits_{n \to \infty } {u_n} = 0\)

If \(\mathop {\lim }\limits_{n \to \infty } {u_n} \ne 0\), the given series is oscillatory.

Calculation:

Given series is \(\mathop \sum \limits_{n = 1}^\infty \frac{{{{\left( { - 1} \right)}^{n - 1}}n}}{{2n - 1}}\)

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,

\(\mathop {\lim }\limits_{n \to \infty } {u_n} = \mathop {\lim }\limits_{n \to \infty } \frac{n}{{2n - 1}} = \frac{1}{2}\)

⇒ \(\mathop {\lim }\limits_{n \to \infty } {u_n} \ne 0\)

∴ 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;

\(\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{u_{n + 1}}}}{{{u_n}}}} \right| = \mathop {{\rm{lim}}}\limits_{n \to \infty } \left| {\left( {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right)} \right|.x\)

If \(\mathop {{\rm{lim}}}\limits_{n \to \infty } \left( {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right) = l,\) 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 \(1 + \frac{2}{5}x + \frac{6}{9}{x^2} + \ldots + \frac{{{2^n} - 2}}{{{2^n} + 1}}{x^{n - 1}} + \ldots (x > 0)\)

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

\(\mathop {\lim }\limits_{n \to \infty } \frac{{{u_n}}}{{{u_{n + 1}}}} = \frac{{1 - 0}}{{1 + 0}}.\frac{{2 + 0}}{{2 - 0}}.\frac{1}{x} = \frac{1}{x}\)

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, \(\mathop {\lim }\limits_{n \to \infty } {u_n} = \mathop {\lim }\limits_{n \to \infty } \frac{{{2^n} - 2}}{{{2^n} + 1}} = \mathop {\lim }\limits_{n \to \infty } \frac{{1 - \frac{2}{{{2^n}}}}}{{1 + \frac{1}{{{2^n}}}}} = 1 \ne 0\)

⇒ ∑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 \(\mathop {\lim }\limits_{n \to \infty } {\left( {{u_n}} \right)^{\frac{1}{n}}} = \lambda \)

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

Calculation:

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

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

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

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

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

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

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

The given alternating series is:

\(x - \frac{{{x^2}}}{{{2^2}}} + \frac{{{x^3}}}{{{3^2}}} - \frac{{{x^4}}}{{{4^2}}} + \ldots \infty \)

\({u_n} = \frac{{{{≤ft( { - 1} }^{n - 1}}{x^n}}}{{{n^2}}}\)

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

\(≤ft| {{u_n}} | = \frac{{{{≤ft| x |}^n}}}{{{n^2}}}\)

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

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

Now, \(\mathop {\lim }\limits_{n \to \infty } \frac{{≤ft| {{u_n}} |}}{{≤ft| {{u_{n + 1}}} |}} = \mathop {\lim }\limits_{n \to \infty } \frac{{{{≤ft( {n + 1} )}^2}}}{{{n^2}}} \times \frac{1}{{≤ft| x |}} = \frac{1}{{≤ft| x |}}\)

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

Absolute series is divergent for \(\frac{1}{{≤ft| x |}} < 1\) 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

\(\mathop \sum \limits_{n = 1}^\infty ≤ft| {{u_n}} | = \mathop \sum \limits_{n = 1}^\infty \frac{1}{{{n^2}}}\)

\(≤ft| {{u_n}} | = \frac{1}{{{n^2}}}\)

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

n < (n+1), for all n

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

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

\(≤ft| {{u_{n + 1}}} | < ≤ft| {{u_n}}|\); for all n

Moreover, \(\mathop {\lim }\limits_{n \to \infty } ≤ft| {{u_n}} | = \mathop {\lim }\limits_{n \to \infty } \frac{1}{{{n^2}}} = 0\)

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

\(\mathop \sum \limits_{n = 1}^\infty {{u_n}} = \mathop \sum \limits_{n = 1}^\infty \frac{-1}{{{n^2}}}\)

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

Concept Used:-

If the summation from 0 to n such that \(\rm\displaystyle\sum_{r = 0}^{ n}\) ar xr is given, then the expanded form of it can be written as,

 ⇒ \(\rm\displaystyle\sum_{r = 0}^{ n}\) ar xr = a₀+a₁ x+a₂ x²+a₃ x³+a₄ x⁴+..........+a_nxⁿ

Explanation:-

Given,

(1+ x + x2)n = \(\rm\displaystyle\sum_{r = 0}^{2 n}\) ar xr, 

On the expanding right-hand side, we get,

\(\left(1+x+x^2\right)^n=a_0+a_1 x+a_2 x^2+a_3 x^3+a_4 x^4+\cdots+a_{2 n} x^{2 n}\)

Differentiate it with respect to x,

\(n\left(1+x+x^2\right)^{n-1}(1+2 x)=a_1+2 a_2 x+3 a_3 x^2+4 a_4 x^3+\cdots+2 n a_n x^{2 x-1}\)Now put x = -1 in the above equation,

\(\Rightarrow n(1-1+1)^{n-1}(1-2)=a_1-2 a_2+3 a_3-4 a_4+5a_5 \cdots-2 n a_{2n}\\ \Rightarrow n^1(1)^{n-1}(-1)=a_1-2 a_2+3 a_3-4 a_4+5a_5 \cdots-2 n a_{2n}\\ \Rightarrow -n=a_1-2 a_2+3 a_3-4 a_4+5a_5 \cdots-2 n a_{2n}\)

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

Hence, the correct option is 3.