Sequences and Series Test

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Sequences and Series Test - 52

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Weekly Quiz Competition

Question 1
1 / -0

If $$n$$ is an odd integer greater than or equal to $$1$$, then the value of $$ n^3-(n-1)^3+(n-2)^3-....+(-1)^{n-1}1^3$$, is

A
$$\dfrac{(n+1)^2(2n-1)}{4}$$

B
$$ \dfrac{(n-1)^2(2n-1)}{4}$$

C
$$\dfrac{(n+1)^2(2n+1)}{4}$$

D
$$\dfrac{(n+1)^2(2n+1)}{8}$$

Solution

The given series is
$$S = 1^3 -2^3 +3^3 -4^3 +5^3 -6^3+...-(n-1)^3+n^3$$
$$\because n$$ is an odd integer.
This is the same as
$$S = (1^3 +2^3 +3^3 +...+n^3)-2(2^3+4^3+6^3+...+(n-1)^3)$$
Which can be written as
$$S = (1^3 +2^3 +3^3 +...+n^3)-16\left(1^3 +2^3 +3^3 +...\left(\dfrac {n-1}{2}\right)^3\right)$$
Which can now be solved using
$$\displaystyle\sum r^3 =\dfrac {n^2(n+1)^2}{4}$$

$$\therefore$$ The desired answer is
$$S = \dfrac{n^2(n+1)^2}{4} - 16\dfrac{\left(\dfrac{n-1}{2}\right)^2\left(\dfrac{n+1}{2}\right)^2}{4} = \dfrac{(n+1)^2(2n-1)}{4}$$

Hence, option A is the correct answer.
Question 2
1 / -0

If $$ \displaystyle f(r)=1+\frac {1}{2}+\frac {1}{3}+.....+\frac {1}{r}$$ and $$f(0)=0$$, then value of $$ \displaystyle \sum_{r=1}^{n}(2r+1)f(r)$$ is

A
$$n^2f(n)$$

B
$$ \displaystyle (n+1)^2f(n+1)-\frac {n^2+3n+2}{2}$$

C
$$ \displaystyle (n+1)^2f(n)-\frac {n^2+n+1}{2}$$

D
$$(n+1)^2f(n)$$

Solution

Since $$ \displaystyle \sum_{r=1}^{n}(2r+1)f(r)=\sum_{r=1}^{n}(r^2+2r+1-r^2)f(r)=\sum_{r=1}^{n}\left \{(r+1)^2-r^2\right \}f(r)$$
$$ \displaystyle =\sum_{r=1}^{n}\left \{(r+1)^2f(r)-(r+1)^2f(r+1)+(r+1)^2f(r+1)-r^2f(r)\right \}$$
$$ \displaystyle =\sum_{r=1}^{n}(r+1)^2\left \{f(r)-f(r+1)\right \}+\sum_{r=1}^{n}\left \{(r+1)^2f(r+1)-r^2f(r)\right \}$$
$$ \displaystyle =-\sum_{r=1}^{n}\frac {(r+1)^2}{(r+1)}+\sum_{r=1}^{n-1}(r+1)^2f(r+1)+(n+1)^2f(n+1)-\sum_{r=1}^{n}r^2f(r)$$
$$ \displaystyle =-\sum_{r=1}^{n}(r+1)+\left \{2^2f(2)+3^2f(3)+ ....+n^2f(n)\right \}+(n+1)^2f(n+1)-\left \{1^2f(1)+2^2f(2)+3^2f(3)+ .....+n^2f(n)\right \}$$
$$\displaystyle =-\sum_{r=1}^{n}r-\sum_{r=1}^{n}1+\left((n+1)^2f(n+1)-1^2f(1)\right)=-\frac {n(n+1)}{2}-n+(n+1)^2f(n+1)-f(1)$$
$$ \displaystyle =(n+1)^2f(n+1)-\frac {n(n+3)}{2}-1 =(n+1)^2f(n+1)-\frac {(n^2+3n+2)}{2}$$
Question 3
1 / -0

The sum to $$n$$ terms of the series $$\displaystyle \frac{3}{1^2}+\frac{5}{1^2+2^2}+\frac{7}{1^2+2^2+3^2}+...$$ is

A
$$\displaystyle \frac{6n}{n+1}$$

B
$$\displaystyle \frac{9n}{n+1}$$

C
$$\displaystyle \frac{12n}{n+1}$$

D
$$\displaystyle \frac{3n}{n+1}$$

Solution

The observation is important ...we can see that the numerator general term is $$2n+1$$

while the denominator is the sum of $$n^2$$ terms so the generalized term

$$Tn=(2n+1)/[n(n+1)(2n+1)/6]=6/[n(n+1)]=6[(1/n)-(1/(n+1))]$$

Therefore when we add five terms we can see all terms will cancel out except the first and the last term.

So $$S=6[1-1/(n+1)]=6n/(n+1)$$

Option A
Question 4
1 / -0

The sum of the first $$n$$ terms of the series $$1^2+2\cdot2^2+3^2+2\cdot4^2+5^2+2\cdot6^2+..... $$ is $$\dfrac{n \left ( n+1\right )^2}{2}$$ when $$n$$ is even.
When $$n$$ is odd, then the sum is

A
$$ \dfrac{3n \left ( n+1\right )}{2}$$

B
$$ \dfrac{n (n+1)^2}{4}$$

C
$$\dfrac{n \left ( n+1\right )^2}{2}$$

D
$$\dfrac{n^2 \left ( n+1\right )}{2}$$

Solution

If $$n$$ is odd, then $$ n-1$$ is even.
$$(n-1)^{th}$$ term is $$2(n-1)^2$$ the $$n^{th}$$ term is $$n^2$$
Hence, using the given formula is valid if $$n$$ is even, so substitute $$n-1$$ in that formula in place of $$n$$ and add $$n^2$$
We get,
Sum $$=\dfrac{(n-1)n^2}{2} + n^2=\dfrac{n^2(n+1)}{2}$$
Question 5
1 / -0

$$\displaystyle \frac{1^{3}}{1}+\frac{1^{3}+2^{3}}{1+3}+\frac{1^{3}+2^{3}+3^{3}}{1+3+5}+\ldots.n$$ terms $$=$$

A
$$\displaystyle \frac{\mathrm{n}(2\mathrm{n}^{2}+9\mathrm{n}+13)}{24}$$

B
$$\displaystyle \frac{n(2n^{3}+9n+13)}{8}$$

C
$$\displaystyle \frac{n(n^{2}+9n+13)}{24}$$

D
$$\displaystyle \frac{n(n^{2}+9n+13)}{8}$$

Solution

General $$\displaystyle k^{th}$$ term $$ = \dfrac{\sum_{i=1}^{k}i^3}{\sum_{i=1}^{k}(2i-1)}$$ $$\displaystyle =\frac{(k+1)^2}{4}$$

Final answer : $$\displaystyle \sum_{k=1}^n\dfrac{(k^2+2k+1)}{4}$$

$$\implies \displaystyle S_n = \dfrac{1}{4}\displaystyle \sum_{k=1}^n k^2+ 2k+1$$

We know,
$$\sum_{i=1}^{n} k =1 + 2 + 3 +… + n = \dfrac{n(n+1)}{2}$$ (sum of first n natural numbers)
$$ \sum_{i=1}^{n} k^2 =1^2 + 2^2 + 3^2 +… + n^2=\dfrac{n(n+1)(2n+1)}{6}$$ (sum of squares of the first n natural numbers)

$$\therefore S_n =\displaystyle \frac{n(n+1)(2n+1)}{24}+\frac{n(n+1)}{4}+\frac{n}{4}$$

$$\displaystyle=n\frac{(2n^2+9n+13)}{24}$$

Hence, option 'A' is correct.
Question 6
1 / -0

Observe the following lists List I and List II

(A) $$\displaystyle \sum_{n=0}^{\infty}\frac{x^{n}(\log_{e}a)^{n}}{n!}$$ $$(1)\displaystyle \frac{e^{x}-e^{-x}}{2}$$
(B) $$\displaystyle \sum_{n=0}^{\infty}\frac{x^{2n}}{(2n)!}$$ $$(2) e^{-ax}$$
(C) $$\displaystyle \sum_{n=0}^{\infty}\frac{x^{2n+1}}{(2n+1)!}=$$ $$(3) a^{x}$$
(D) $$\displaystyle \sum_{n=0}^{\infty}\frac{(-1)^{n}.(ax)^{n}}{n!}$$ $$(4)\displaystyle \frac{a^{x}-a^{-x}}{2}$$
$$(5)\displaystyle \frac{e^{x}+e^{-x}}{2}$$
The correct match of List I to List II is:

A
A - 1, B - 4, C - 3, D - 5

B
A - 4, B - 2, C - 1, D - 3

C
A - 3, B - 5, C - 1, D - 2

D
A - 2, B - 3, C - 5, D - 4

Solution

(A) When we expand it, we will get

$$1+\dfrac { { x }^{ }{ \log _{ e }{ a } }^{ } }{ 1 } + \dfrac { { x }^{ 2 }{( \log _{ e }{ a } })^{2 } }{ 2! }+....$$ This is a general expression for the expansion of $${ a }^{ x }$$
So, A - 3
(B) Expansion is given by $$1+\dfrac { { x }^{ 2 } }{ 2! } +\dfrac { { x }^{ 4 } }{ 4! } +\dfrac { { x }^{ 6 } }{ 6! } ..$$ which is $$\dfrac { { e }^{ x }+{ e }^{ -x } }{ 2 } $$
So, B - 5

(C) Expansion is given by $$1+\dfrac { { x }^{ 3 } }{ 3! } +\dfrac { { x }^{ 5 } }{ 5! } +\dfrac { { x }^{ 7 } }{ 7! } ..$$ and this is $$\dfrac { { e }^{ x }-{ e }^{ -x } }{ 2 } $$
So, C - 1

(D) Expansion is given by $$1-\dfrac { ax }{ 1! } +\dfrac { (ax)^{ 2 } }{ 2! } -\dfrac {(ax)^{3}}{ 3! }....$$ which is an expansion of $${ e }^{ -ax }$$
So, D - 2
Question 7
1 / -0

Sum the series $$1^3+3^3+5^3+.......... $$ to $$n$$ terms is

A
$$ n^2\left ( 2n^2-1 \right )$$

B
$$n\left ( 2n^2-1 \right )$$

C
$$n\left ( 2n^2+1 \right )$$

D
$$n^2\left ( 2n^2+1 \right )$$

Solution

$$n^{th}$$ term of the series will be $$(2n-1)^3$$

$$1^3 +3^3 +5^3+.....+(2n-1)^3 = 1^3 +2^3 +3^3+4^3+.....(2n)^3 -(2^3 +4^3+6^3+......(2n)^3)$$

$$=\left (\dfrac{2n(2n+1)}{2}\right)^2-2^3(1^3+2^3+3^3+.....+n^3)$$

$$=n^2(2n+1)^2 - 2n^2(n+1)^2$$

$$=4n^4+4n^3+n^2-2n^4-4n^3-2n^2$$

$$=2n^4-n^2$$

$$=n^2(2n^2-1)$$
Question 8
1 / -0

ABCD is a square of length a, $$a\epsilon N$$, a>1. Let $$L_{1}$$, $$L_{2}$$, $$L_{3}$$,... be points on BC such that $$BL_{1}=L_{1}L_{2}=L_{2}L_{3}=...=1$$ and $$M_{1}$$, $$M_{2}$$, $$M_{3}$$,... be points on CD such that $$CM_{1}=M_{1}M_{2}=M_{2}M_{3}=...=1$$. Then $$\sum_{n=1}^{a-1}\left ( A{L_{n}}^{2}+L_{n}{M_{n}}^{2} \right )$$ is equal to

A
$$\dfrac{1}{2}a\left ( a-1 \right )^{2}$$

B
$$\dfrac{1}{2}a\left ( a-1 \right )\left ( 4a-1 \right )$$

C
$$\dfrac{1}{2}a\left ( a-1 \right )\left ( 2a-1 \right )\left ( 4a-1 \right )$$

D
none of these

Solution

$$A{ { L }_{ n } }^{ 2 }={ a }^{ 2 }+{ n }^{ 2 }\\ { { L }_{ n }{ M }_{ n } }^{ 2 }={ (a-n) }^{ 2 }+{ n }^{ 2 }\\ A{ { L }_{ n } }^{ 2 }+{ { L }_{ n }{ M }_{ n } }^{ 2 }=2{ a }^{ 2 }+{ 3n }^{ 2 }-2na$$
Thus, the summation is:
$$2{ a }^{ 2 }(a-1)+3.\dfrac { (a-1)(a)(2a-1) }{ 6 } -2a.\dfrac { (a-1)(a) }{ 2 } \\ =\dfrac { a }{ 2 } .(a-1).\left( 4a+2a-1-2a \right) =\dfrac { 1 }{ 2 } a.(a-1).\left( 4a-1 \right) $$
Question 9
1 / -0

The sum of n terms of the series $$1 + (1 + a) + (1 + a + a^2) + (1 + a + a^2 + a^3) + ....$$ is

A
$$\displaystyle \frac{n}{1+a} - \frac{1 - a^n}{(1 - a)^2}$$

B
$$\displaystyle \frac{n}{1- a} + \frac{a (1 - a^n)}{(1 - a)^2}$$

C
$$\displaystyle \frac{n}{1 - a} + \frac{a (1 + a^n)}{(1 - a)^2}$$

D
none of the above
Question 10
1 / -0

If $$\displaystyle \sum_{r=1}^{n}t_{r}=\frac {n(n+1)(n+2)(n+3)}{8}$$, then $$\displaystyle \underset{n\rightarrow \infty}{ \lim} \sum_{r=1}^{n}\frac {1}{t_{r}}$$
is equal to:

A
$$\displaystyle \frac {1}{8}$$

B
$$\displaystyle \frac {1}{4}$$

C
$$\displaystyle \frac {1}{2}$$

D
$$1$$

Solution

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Sequences and Series Test - 52

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Sequences and Series Test - 52

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Question 1

$$\dfrac{(n+1)^2(2n-1)}{4}$$

$$ \dfrac{(n-1)^2(2n-1)}{4}$$

$$\dfrac{(n+1)^2(2n+1)}{4}$$

$$\dfrac{(n+1)^2(2n+1)}{8}$$

Solution

Question 2

$$n^2f(n)$$

$$ \displaystyle (n+1)^2f(n+1)-\frac {n^2+3n+2}{2}$$

$$ \displaystyle (n+1)^2f(n)-\frac {n^2+n+1}{2}$$

$$(n+1)^2f(n)$$

Solution

Question 3

$$\displaystyle \frac{6n}{n+1}$$

$$\displaystyle \frac{9n}{n+1}$$

$$\displaystyle \frac{12n}{n+1}$$

$$\displaystyle \frac{3n}{n+1}$$

Solution

Question 4

$$ \dfrac{3n \left ( n+1\right )}{2}$$

$$ \dfrac{n (n+1)^2}{4}$$

$$\dfrac{n \left ( n+1\right )^2}{2}$$

$$\dfrac{n^2 \left ( n+1\right )}{2}$$

Solution

Question 5

$$\displaystyle \frac{\mathrm{n}(2\mathrm{n}^{2}+9\mathrm{n}+13)}{24}$$

$$\displaystyle \frac{n(2n^{3}+9n+13)}{8}$$

$$\displaystyle \frac{n(n^{2}+9n+13)}{24}$$

$$\displaystyle \frac{n(n^{2}+9n+13)}{8}$$

Solution

Question 6

A - 1, B - 4, C - 3, D - 5

A - 4, B - 2, C - 1, D - 3

A - 3, B - 5, C - 1, D - 2

A - 2, B - 3, C - 5, D - 4

Solution

Question 7

$$ n^2\left ( 2n^2-1 \right )$$

$$n\left ( 2n^2-1 \right )$$

$$n\left ( 2n^2+1 \right )$$

$$n^2\left ( 2n^2+1 \right )$$

Solution

Question 8

$$\dfrac{1}{2}a\left ( a-1 \right )^{2}$$

$$\dfrac{1}{2}a\left ( a-1 \right )\left ( 4a-1 \right )$$

$$\dfrac{1}{2}a\left ( a-1 \right )\left ( 2a-1 \right )\left ( 4a-1 \right )$$

none of these

Solution

Question 9

$$\displaystyle \frac{n}{1+a} - \frac{1 - a^n}{(1 - a)^2}$$

$$\displaystyle \frac{n}{1- a} + \frac{a (1 - a^n)}{(1 - a)^2}$$

$$\displaystyle \frac{n}{1 - a} + \frac{a (1 + a^n)}{(1 - a)^2}$$

none of the above

Question 10

$$\displaystyle \frac {1}{8}$$

$$\displaystyle \frac {1}{4}$$

$$\displaystyle \frac {1}{2}$$

$$1$$

Solution

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