Special Series MCQ Quiz in मराठी - Objective Question with Answer for Special Series - मोफत PDF डाउनलोड करा

Calculations:

Consider, \(\rm \frac{1}{{1 \times 2}} + \frac{1}{{2 \times 3}} + \frac{1}{{3 \times 4}} +...+\frac{1}{{99 \times 100}} \) = \(\rm \sum _{n =1}^{100} \dfrac {1}{n\times (n+1)}\) 

\(\rm \dfrac {1}{n \times (n+1)} = \dfrac {1}{n} - \dfrac {1}{(n+1)}\)

⇒\(\rm \sum_{n = 1}^{100} \dfrac {1}{n(n+1)} = \sum_{n = 1}^{100} [\dfrac {1}{n} - \dfrac {1}{(n+1)}]\)

⇒\(\rm \sum_{n = 1}^{100} \dfrac {1}{n(n+1)} = (1 - \dfrac 12)+ ( \dfrac 12 - \dfrac 13)+ ( \dfrac 13- \dfrac 14)+ ....( \dfrac 1{99}- \dfrac 1{100})\)

⇒\(\rm \sum_{n = 1}^{100} \dfrac {1}{n(n+1)} = (1 - \dfrac 1{100})\)

Hence, 

Formula Used: 

Sum of square of n numbers = \(\frac{n(n +1)(2n + 1)}{6}\)

Sum of n terms = \(\frac{n(n +1)}{2}\)

Calculation:      

a₁ = 1

a₂ = 1 + 8 = 9

a₃ = 9 + 8 + 7 = 24

a₄ = 24 + 8 + 7 + 7 =  46

So, then K^th term will be:

⇒ a_k = 1 + 8(k - 1) + 7\(\frac{(k -1)(k - 2)}{2}\)

⇒ a_k = 7k²/2 - 5k/2

Sum of first n term:

\(⇒ \frac{7}{2} \times \frac{1}{6}n(n +1)(2n + 1) - \frac{5}{2} \times \frac{1}{2}n(n + 1)\)

\(⇒ \frac{1}{12}n(n + 1).[7(2n + 1) - 15]\)

\(⇒ \frac{1}{12}n(n + 1).(14n - 8)\)

\(∴ \frac{1}{6}n(n + 1).(7n - 4)\)

The correct option is 4 i.e. \(\frac{1}{6}n(n + 1).(7n - 4)\)   

Concept:

Sum of first 'n' natural numbers = \(\rm \sum_{1}^{n} r = \dfrac{n(n + 1)}{2}\)

Sum of the square of first n natural numbers = \(\rm \sum_{1}^{n} r^2= \dfrac{n(n + 1)(2n + 1)}{6}\)

Sum of cubes of first n natural numbers = \(\rm \sum_{1}^{n} r^3 = \left [\rm\frac{n(n + 1)}{2} \right]^2\)

Calculation:

To Find: n^th term of the series 

\(\frac {1^3}{1} + \frac {1^3 + 3^3}{1 + 3} + \frac {1^3 + 3^3 + 5^3}{1 + 3 + 5}+...\)

\(\rm T_n = \dfrac{1^3+3^3+5^3+.... \text{upto n terms}}{1+3+5+....\text{upto n terms}}\\=\dfrac{\sum_{1}^{n}(2r+1)^3}{\sum_{1}^{n}(2r+1)}\\=\dfrac{\sum_{1}^{n}(8r^3+12r^2+6r+1)}{\sum_{1}^{n}(2r+1)}\\=\dfrac{\sum_{1}^{n}8r^3+\sum_{1}^{n}12r^2+\sum_{1}^{n}6r+\sum_{1}^{n}1}{\sum_{1}^{n}2r+\sum_{1}^{n}1}\)

\(\rm =\dfrac{8\sum_{1}^{n}r^3+12\sum_{1}^{n}r^2+6\sum_{1}^{n}r+\sum_{1}^{n}1}{2\sum_{1}^{n}r+\sum_{1}^{n}1}\\\rm = \dfrac{8 \times \left [\rm\frac{n(n + 1)}{2} \right]^2 + 12 \times \frac{n(n + 1)(2n + 1)}{6} + 6 \times \frac{n(n + 1)}{2}+n}{2 \times \frac{n(n + 1)}{2}+n}\\=\dfrac{2n^2(n+1)^2 +2n(n+1)(2n+1)+3n(n+1)+n}{n(n+1)+n}\)

\(\rm =\dfrac{2n(n+1)^2 +2(n+1)(2n+1)+3(n+1)+1}{(n+1)+1}\\=\dfrac{(n+1)[2n^2+2n+4n+2+3]+1}{n+2}\\=\dfrac{(n+1)[2n^2+6n+5]+1}{n+2}\)

Concept:

• Sum of the first n Natural Numbers \(\rm 1+2+3+4+....+n =\sum n = \frac{n(n+1)}{2}\)
• Sum of the Square of the first n Natural Numbers \(\rm 1^2+2^2+3^2+4^2+....+ n^2=\sum n^2 = \frac{n(n+1)(2n+1)}{6}\)
• Sum of the Cubes of the first n Natural Numbers \(\rm 1^3+2^3+3^3+4^3+....+ n^3=\sum n^3 = \frac{[n(n+1)]^2}{4}\)

Calculation:

Here, we have to find the sum of the series 1⋅ 2 + 2 ⋅ 3 + 3 ⋅ 4 + 4 ⋅ 5 +.....+ 8 ⋅ 9

As we know that, the nth term of the given series is given by T_n = n (n + 1)

Sum of series = \(\rm S_n = \sum T_n= \sum (n^2+n)=\sum n^2+\sum n\)

As we know that, \(\rm 1^2+2^2+3^2+4^2+....+ n^2=\sum n^2 = \frac{n(n+1)(2n+1)}{6}\) and \(\rm 1+2+3+4+....+n =\sum n = \frac{n(n+1)}{2}\)

\(⇒ \rm S_n = \frac{n(n+1)(n+2)}{3}\)

Here, n = 8

\(⇒ \rm S_8 = \frac{8 \times 9 \times 10}{3} = 240\)

Hence, option 1 is the correct answer.

Concept:

The general term of a sequence a_n = S_{n + 1} - S_n,

Where S_n is the sum of first n terms of the series and S_{n + 1} is the sum of first (n + 1) terms of the series.

Calculation:

Given: S_n = n + 12

By substituting n by (n + 1) in the above equation, we get

S_{n + 1} = (n + 1) + 12 = n + 13

Now, General term of a sequence

⇒ a_n = S_{n + 1} - S_n = (n + 13) - (n + 12) = 1 ∀ n ∈ N

⇒ an = 1

∴ a₃ = 1

S₁ = a₁ = 1 + 12 = 13

We can write the series as 13, 1, 1, 1, 1, ............

Concept:

If a1, a2, a3,...are in GP with common ratio r, 

 \(​​​​r=\frac{a_{2}}{a_{1}}=\frac{a_{3}}{a_{2}}=\frac{a_{4}}{a_{3}}\)

If a be the first term, r be the common ratio of a GP then,

 \(\rm S_n = \frac{a(1-r^n)}{1-r}\)

Calculation:

Calculation:

Let required sum is S.

⇒ S = \(\frac{1}{2} + \frac{3}{4}+\frac{7}{8}+\frac{15}{16}+...\)

⇒ S =\(\frac{2-1}{2} + \frac{4-1}{4}+\frac{8-1}{8}+\frac{16-1}{16}+...\)

⇒ S = \(1-\frac{1}{2} +1- \frac{1}{4}+1-\frac{1}{8}+1-\frac{1}{16}+...n\ term\)

⇒ S = \( n-[\frac{1}{2} + \frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...n\ term]\)

Threfore, a = 1/2 and r = 1/2

We know that the sum of the n^th term of GP (r < 1) is 

\(\rm S_n = \frac{a(1-r^n)}{1-r}\)

⇒ S = \(n- \frac{(1/2)[1-(1/2)^n]}{1-1/2}\)

⇒ S = \(n- \frac{(1/2)[1-(2)^{-n}]}{1/2}\)          (∵ 1/aⁿ = a^-n)

⇒ S = n - 1 + 2-n

∴ S = n + 2-n -1

Given:

\(\dfrac{1}{1.2.3}+\dfrac{3}{2.3.4}+\dfrac{5}{3.4.5}+\dfrac{7}{4.5.6}+...\)

Calculations:

Solving the given series,

\(\Rightarrow \displaystyle\sum_{r=1}^n \frac{2r - 1}{r(r+1)(r+2)} \)

\(\Rightarrow \displaystyle\sum_{r=1}^n \frac{2}{(r+1)(r+2)} - \frac{1}{r(r+1)(r+2)}\)

\(\Rightarrow 2\displaystyle\sum_{r=1}^n \frac{(r+2)-(r+1)}{(r+1)(r+2)} - \frac{1}{2}\left(\displaystyle\sum_{r=1}^n \frac{1}{r(r+1)}-\frac{1}{(r+1)(r+2)}\right) \)

\(\Rightarrow 2\displaystyle\sum_{r=1}^n \left(\frac{1}{r+1}-\frac{1}{r+2}\right) - \frac{1}{2}\left(\displaystyle\sum_{r=1}^n \frac{1}{r}-\frac{1}{r+1}\right) +\frac{1}{2}\displaystyle\sum_{r=1}^n \left(\frac{1}{r+1}-\frac{1}{r+2}\right)\)

\(\Rightarrow 2\left(\dfrac{1}{2} -\dfrac{1}{n+2}\right)-\dfrac{1}{2}\left(1 -\dfrac{1}{n+1}\right)+\dfrac{1}{2}\left(\dfrac{1}{2} -\dfrac{1}{n+2}\right)\)

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

After simplifying,

\(\Rightarrow \dfrac{n(3n+1)}{4(n+1)(n+2)}\)

∴ The sum of 'n' terms of the series is \(\dfrac{n(3n+1)}{4(n+1)(n+2)}\)

Given:

We need to find the sum of the squares of all two-digit numbers which are completely divisible by 4.

Formula Used:

Sum of squares = \(\sum_{i=1}^{n} (\text{number})^2\)

Calculation:

Two-digit numbers divisible by 4 are: 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96

Sum of squares = 12² + 16² + 20² + 24² + 28² + 32² + 36² + 40² + 44² + 48² + 52² + 56² + 60² + 64² + 68² + 72² + 76² + 80² + 84² + 88² + 92² + 96²

⇒ Sum of squares = 144 + 256 + 400 + 576 + 784 + 1024 + 1296 + 1600 + 1936 + 2304 + 2704 + 3136 + 3600 + 4096 + 4624 + 5184 + 5776 + 6400 + 7056 + 7744 + 8464 + 9216

⇒ Sum of squares = 78320

The sum of the squares of all two-digit numbers completely divisible by 4 is 78320.

Concept:

• Sum of the first n Natural Numbers \(\rm 1+2+3+4+....+n =\sum n = \frac{n(n+1)}{2}\)
• Sum of the Square of the first n Natural Numbers \(\rm 1^2+2^2+3^2+4^2+....+ n^2=\sum n^2 = \frac{n(n+1)(2n+1)}{6}\)
• Sum of the Cubes of the first n Natural Numbers \(\rm 1^3+2^3+3^3+4^3+....+ n^3=\sum n^3 = \frac{[n(n+1)]^2}{4}\)

Calculation:

Here, we have to find the sum of the series 12 ⋅ 2 + 22⋅3 + 32⋅4 + 42⋅5 +.....+up to n terms

nth term of the given series is T_n = n² ⋅ (n + 1) = n3 + n² 

Sum of n terms of the series,  \(\rm S_n = \sum T_n= \sum (n^3+n^2)=\sum n^3+\sum n^2\)

As we know that, 

\(\rm 1^2+2^2+3^2+4^2+....+ n^2=\sum n^2 = \frac{n(n+1)(2n+1)}{6}\) and \(\rm 1^3+2^3+3^3+4^3+....+ n^3=\sum n^3 = \frac{[n(n+1)]^2}{4}\)

\(\Rightarrow \rm S_n =\left ( { \frac{n(n+1)}{2}} \right )^2+\frac{n(n+1)(2n+1)}{6}\)

\(\Rightarrow \rm S_n =\left ( { \frac{n(n+1)}{2}} \right )\left ( \frac{n(n+1)}{2}+\frac{2n +1}{3} \right )\)

\(\Rightarrow \rm S_n =\left ( { \frac{n(n+1)}{2}} \right )\left ( \frac{3n^2+7n+2}{6} \right )\)

\(\Rightarrow \rm S_n =\left ( { \frac{n(n+1)}{2}} \right )\left ( \frac{(3n+1)(n+2)}{6} \right )\)

\(\Rightarrow \rm S_n =\left ( { \frac{n(n+1)(3n+1)(n+2)}{12}} \right )\)

Hence, option 2 is the correct answer.

Concept:

• Sum of the first n Natural Numbers \(\rm 1+2+3+4+....+n = \frac{n(n+1)}{2}=\sum n\)
• Sum of the Square of the first n Natural Numbers \(\rm 1^2+2^2+3^2+4^2+....+ n^2=\sum n^2 = \frac{n(n+1)(2n+1)}{6}\)
• Sum of the Cubes of the first n Natural Numbers \(\rm 1^3+2^3+3^3+4^3+....+ n^3=\sum n^3 = \frac{[n(n+1)]^2}{4}\)

Calculation:

Here, we have to find the sum of the series \(\rm \frac{1}{4}+\frac{1}{2}+\frac{3}{4}+\frac{1}{1}+\frac{5}{4}+....+upto\ 16 \ terms\)

Let S_n = \(\rm \frac{1}{4}+\frac{1}{2}+\frac{3}{4}+\frac{1}{1}+\frac{5}{4}+.... + \frac{n}{4}\)

Multiplying the above equation by 4 we get,

⇒ 4S_n = 1 + 2 + 3 + 4 + 5 +.....+ n

As we know that, \(\rm 1+2+3+4+....+n = \frac{n(n+1)}{2}=\sum n\)

⇒ 4S_n = n(n + 1)/2

⇒ S_n = n(n + 1)/8

Now substitute n = 16 in the above equation we get,

⇒ S₁₆ = 17 × 2 = 34

Hence, option 4 is the correct answer.