Sequences and Series Test

Result

Sequences and Series Test - 18

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

The sum of the series $$1^2 - 2^2 + 3^2 - 4^2 + .......... - 1000^2 + 1001^2$$ is

A
$$-501\times1001$$

B
$$501\times1001$$

C
$$499\times1001$$

D
$$-499\times1001$$

Solution

We would like to regroup the series as follows:
$$ 1^2+(3^2-2^2)+(5^2-4^2)+....(1001^2-1000^2) $$
$$a^2-b^2=(a+b)(a-b)$$ , in our case $$a-b=1$$
Hence,
$$1+5+9+13+... +2001$$
There are $$501$$ terms, then
Sum$$=501\times1001$$
Question 2
1 / -0

$$ 1.4+2.7+3.10+...+\mathrm{n}(3\mathrm{n}+1)=$$

A
$$n(2n+1)^{2}$$

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

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

D
$$(\mathrm{n}+2)(\mathrm{n}+3)$$

Solution

$$\sum _{ k=1 }^{ n }{ n(3n+1) }= \sum _{ k=1 }^{ n }{ 3n^2+n }=3\sum _{ k=1 }^{ n }{n^2 }+\sum _{ k=1 }^{ n }{ n }$$

$$=3\times \dfrac{n(n+1)(2n+1)}{6} + \dfrac{n(n+1)}{2}$$

$$ = \cfrac{n(n+1)(2n+1)}2 + \cfrac{n(n+1)}2$$

$$=n(n+1)^2$$
Question 3
1 / -0

The sum of $$7$$ terms of the series $$1^2-2^2+3^2-4^2+5^2-6^2+...$$ is

A
$$-21$$

B
$$15$$

C
$$28$$

D
$$-35$$

Solution

We can see that first 6 terms taken in pair and is equal to $$-3+(-7)+(-11)$$
So the sum will be $$-21+7^2$$
$$=28$$
Question 4
1 / -0

$$1^{2}+1+2^{2}+2+3^{2}+3+\ldots..+n^{2}+n=$$

A
$$\displaystyle \frac{(n+1)(n+2)}{3}$$

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

C
$$\displaystyle \frac{n(n+2)(n+3)}{6}$$

D
$$\displaystyle \frac{(\mathrm{n}+1)(\mathrm{n}+2)(\mathrm{n}+3)}{4}$$

Solution

$$S=\Sigma n^2 + \Sigma n $$
Therefore,
$$S=\dfrac{n(n+1)(2n+1)}6+\dfrac{n(n+1)}2 $$
Hence,
$$S=\dfrac{n(n+1)(n+2)}3 $$
Question 5
1 / -0

$$\displaystyle \frac{1.2^{2}+2.3^{2}+3.4^{2}+.\cdot.\cdot.\cdot.\cdot+n(n+1)^{2}}{1^{2}.2+2^{2}.3+3^{2}.4++n^{2}(n+1)}=$$

A
$$\displaystyle \frac{3n+1}{3n+5}$$

B
$$\displaystyle \frac{3n+5}{3n+1}$$

C
$$(3n+1)(3n+5)$$

D
$$\displaystyle \frac{3n+5}{3n+7}$$

Solution

Numerator : $$\sum_{k=1}^n k(k+1)^2$$ $$ = \sum_{k=1}^n (k^3+2k^2+k) $$
Denominator : $$\sum_{k=1}^n k^2(k+1)$$ = $$ \sum_{k=1}^n (k^3 + k^2) $$
On evaluating each sum we get
Numerator = $$n(n+1)(\frac{n(n+1)}{4} + \frac{2n+1}{3} + \frac{1}{2}) $$
Denominator = $$n(n+1)(\frac{n(n+1)}{4}+\frac{2n+1}{6}) $$
Now,
$$Answer=\frac{numerator}{denominator}=\frac{n(n+1)}{n(n+1)}\times\frac{(n+2)(3n+5)}{(n+2)(3n+1)}=\frac{3n+5}{3n+1}$$
Hence, option 'B' is correct.
Question 6
1 / -0

$$1^3+1^2+1+2^{3}+2^{2}+2+3^{3}+3^{2}+3+ ... 3\mathrm{n} $$ terms $$=$$

A
$$\displaystyle \frac{n(n+1)(n^{2}+12n+5)}{12}$$

B
$$\displaystyle \frac{n(n+1)(3n^{2}+7n+8)}{12}$$

C
$$\displaystyle \frac{n(n+1)(n+2)(n^{2}+5n+6)}{12}$$

D
$$\displaystyle \frac{(\mathrm{n}+1)(\mathrm{n}+2)(\mathrm{n}+3)}{4}$$

Solution

$$S=\Sigma r^3 +\Sigma r^2 +\Sigma r$$

$$S=\dfrac{n^2(n+1)^2}4 +\dfrac{n(n+1)(2n+1)}6 +\dfrac{n(n+1)}2$$

$$S=\dfrac{n(n+1)(3n^2 +7n+8)}{12}$$

Hence Option B
Question 7
1 / -0

The value of $$\sum_{r=1}^{10}\left ( 2^{r-1}+8r-3 \right )$$ is equal to

A
1343

B
1234

C
1334

D
1433

E
1534

Solution

$$\sum_{r=1}^{10}\left ( 2^{r-1}+8r-3 \right )=\sum_{r=1}^{10}2^{r-1}+\sum_{r=1}^{10}\left ( 8r \right )-\sum_{r=1}^{10}3$$

$$=\frac{2^{10}-1}{2-1}+\frac{8.10.11}{2}-30=1023+440-30=1433$$
Question 8
1 / -0

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

A
$$\displaystyle \frac{n(n+3)}{4}$$

B
$$\displaystyle \frac{n(n+3)}{5}$$

C
$$\displaystyle \frac{n(n+2)}{3}$$

D
$$\displaystyle \frac{n(n+5)}{6}$$

Solution

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)

$$\therefore t_{r}=\dfrac {\Sigma k^2}{\Sigma k}=\dfrac {\dfrac {n(n+1)(2n+1)}{6}}{\dfrac {n(n+1)}{2}}=\dfrac {2n+1}{3}$$

$$S=\dfrac{2}{3} \Sigma n +\dfrac{1}{3} \Sigma 1$$

$$S=\dfrac{2}{3} {\dfrac {n(n+1)}{2}} +\dfrac{1}{3} n$$

$$S=\dfrac {n(n+2)}{3}$$
Question 9
1 / -0

Find $$\displaystyle \sum_{n=1}^{\infty}\frac{1}{(2n-1)!}$$

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

B
$$\displaystyle \frac{1}{2}(e-e^{-1})$$

C
$$e+e^{-1}$$

D
$$\mathrm{e}-\mathrm{e}^{-1}$$

Solution
Question 10
1 / -0

Evaluate $$1.6+2.9+3.12+...+\mathrm{n}(3\mathrm{n}+3)=$$

A
$$\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)$$

B
$$(\mathrm{n}+1)(\mathrm{n}+2)(\mathrm{n}+3)$$

C
$$(\mathrm{n}+2)(\mathrm{n}+3)(\mathrm{n}+4)$$

D
$$(n-1) n(n+1)$$

Solution

$$\sum _{ k=1 }^{ n }{k(3k+3)}=3\sum _{ k=1 }^{ n }{k^2} +3\sum _{ k=1 }^{ n }{k}$$
$$=3\times \dfrac{n(n+1)(2n+1)}{6}+3\times \dfrac{n(n+1)}{2}$$
$$=\dfrac{n(n+1)}{2}(2n+1+3)$$
$$=n(n+1)(n+2)$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Sequences and Series Test - 18

Result

Sequences and Series Test - 18

Score

-

Rank

-

SHARING IS CARING

Question 1

$$-501\times1001$$

$$501\times1001$$

$$499\times1001$$

$$-499\times1001$$

Solution

Question 2

$$n(2n+1)^{2}$$

$$2\mathrm{n}(\mathrm{2n}+1)^{2}$$

$$\mathrm{n}(\mathrm{n}+1)^{2}$$

$$(\mathrm{n}+2)(\mathrm{n}+3)$$

Solution

Question 3

$$-21$$

$$15$$

$$28$$

$$-35$$

Solution

Question 4

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

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

$$\displaystyle \frac{n(n+2)(n+3)}{6}$$

$$\displaystyle \frac{(\mathrm{n}+1)(\mathrm{n}+2)(\mathrm{n}+3)}{4}$$

Solution

Question 5

$$\displaystyle \frac{3n+1}{3n+5}$$

$$\displaystyle \frac{3n+5}{3n+1}$$

$$(3n+1)(3n+5)$$

$$\displaystyle \frac{3n+5}{3n+7}$$

Solution

Question 6

$$\displaystyle \frac{n(n+1)(n^{2}+12n+5)}{12}$$

$$\displaystyle \frac{n(n+1)(3n^{2}+7n+8)}{12}$$

$$\displaystyle \frac{n(n+1)(n+2)(n^{2}+5n+6)}{12}$$

$$\displaystyle \frac{(\mathrm{n}+1)(\mathrm{n}+2)(\mathrm{n}+3)}{4}$$

Solution

Question 7

1343

1234

1334

1433

1534

Solution

Question 8

$$\displaystyle \frac{n(n+3)}{4}$$

$$\displaystyle \frac{n(n+3)}{5}$$

$$\displaystyle \frac{n(n+2)}{3}$$

$$\displaystyle \frac{n(n+5)}{6}$$

Solution

Question 9

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

$$\displaystyle \frac{1}{2}(e-e^{-1})$$

$$e+e^{-1}$$

$$\mathrm{e}-\mathrm{e}^{-1}$$

Solution

Question 10

$$\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)$$

$$(\mathrm{n}+1)(\mathrm{n}+2)(\mathrm{n}+3)$$

$$(\mathrm{n}+2)(\mathrm{n}+3)(\mathrm{n}+4)$$

$$(n-1) n(n+1)$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.