Test: Summation of Series (June 19) - JEE MCQ

Formula Used:

Sum of square of n numbers = 

Calculation:   

Case: 1

1² + 2² + 3² + . . . . . . + 20²

Sum of square of 20 numbers = 

⇒ Sum of square of 20 numbers = 

⇒ Sum of square of 20 numbers = 2870

Case: 2

 1² + 2² + 3² + . . . . . . + 10²

⇒ Sum of square of 10 numbers = 

⇒ Sum of square of 10 numbers = 

⇒ Sum of square of 10 numbers = 385

According to the question: 

⇒ 112 + 122 + 132 + . . . . . . + 202 = Sum of square of 20 numbers - Sum of square of 10 numbers 

⇒ 112 + 122 + 132 + . . . . . . + 202 = 2870 - 385 

∴ 112 + 122 + 132 + . . . . . . + 202 = 2485

The correct option is 4 i.e. 2485

Given: 

series : 1 + 2 + 3 + 4 + 5 + ... + 89 + 90

Formula Used 

Sum of first n natural number = n(n + 1)/2

Calculation 

1 + 2 + 3 + 4 + 5 + ... + 89 + 90

For the given series, n = 90

So, Sum = (90 × 91)/2 = 45 × 91 = 4095 

∴ The sum  of series is 4095

Given:

The numbers are even and lie between 0 and 51

Formula used;

Numbers of terms in AP (n) = [(l - a)/d + 1]

Sum of AP = (n/2) × (a + l)

Here a is first term, l is last term, d is common difference

Calculation:

The numbers are 2, 4, 6, ..... 50

Total even numbers are [(50 - 2)/2 + 1] = 25

Sum of terms = (25/2) × (2 + 50) = 25 × 26

⇒ Sum of terms = 650

∴ 650 is the correct answer.

Given:

Number of terms = 20 

The series is in AP, where 5 + 9 + 13 + 17 + ......

Formulas used:

S_n = n/2[2a + (n - 1)d]

Calculation:

n = 20 

a = 5

d = 9 - 5 = 4 

S₂₀ = 20/2 [ 2 × 5 + 19 × 4]

⇒ 10 [ 10 + 76] 

⇒ 860

∴ The required sum = 860 

Formula used:

S_n = [n x (a + a_n) ] /2

a_n = a + (n-1)d

d = difference

a = initial term

a_n = last term

n = number of terms

S_n = Sum of n terms

Solution:

The series can be written as:

 1/99 [99x99+11 + 99x99+13 + ... + 99x99+67]

= 1/99 [9812 + 9814 + 9816+ ... + 9868]

 Now, our series is, 9812, 9814,...,9868.

a = 9812

a_n = 9868

d = 9814 - 9812 = 2

9868= 9812 + (n-1) x 2

n - 1 = 56/2 = 28

n = 29

S_n = 29 x (9812 + 9868) / 2 = (29 x 19680)/2 = 570720/2 = 285360

Hence, the sum of the series = 285360/99 = 95120/33

Given:

The sum of the first 100 terms of the progression 31 + 34 + 37 + …

Formula Used:

Sum of n numbers  = n/2 × [2a + (n - 1)d]

where: "n" is the number of terms, "a" is the first term, and "d" is the common difference

Calculation:

According to the question, n = 100, a = 31, d = 3

Sum of n numbers  = n/2 × [2a + (n - 1)d]

⇒ 100/2 ×  [2 × 31 + (100-1) × 3]

⇒ 50 × [62 + 297]

⇒ 50 × 359 = 17950

∴ The sum of the first 100 terms of the progression is 17950.

Given:

Numbers = 5 to 14

Concept Used:

Sum of Cubes of n natural numbers = 

Calculation:

⇒ Numbers = 5, 6, 7, 8, 9, 10, 11, 12, 13, 14

⇒ Firstly, we will calculate the sum of the cubes of numbers from 1 to 14,

⇒ Sum of the cubes of first 14 natural numbers = 

⇒ Sum of the cubes of first 4 natural numbers =  = 100

⇒ Sum of the cubes of natural numbers from 5 to 14 = 11025 – 100 = 10,925

Therefore, the sum of the cubes of the natural numbers from 5 to 14 is 10,925.

Given:

1² + 2² + 3² + ... + x² = 

Calculation:

 1² + 2² + 3² + ... + x² = 

Now, taken the 

1² + 2² + 3² + 4² + ... + 19²

Here, total number of terms are 19

⇒ {x(x +1)(2x + 1)}/6

⇒ {19 × (20) × (39)}/6

⇒ 2470      ----(1)

Now, again taken a series of even square number,

2²+ 4² + 6² + ..... + 18²

⇒ 4 + 16 + 36 + ..... + 324

⇒ 4 × (1 + 4 + 9 + ..... + 81)

⇒ 4 × (1² + 2² + 3² + 4² + ... + 9²)

Then,

⇒ {x(x +1)(2x + 1)}/6

Here total number of term are 9,

⇒ {4 × (9 × 10 × 19)}/6

⇒ 1140      ----(2)

equation (1) - equation (2)

⇒ 1² + 3² + 5² + ... + 19² = (1² + 2² + 3² + 4² + ... + 19²) - (2²+ 4² + 6² + ..... + 18²)

⇒  1² + 3² + 5² + ... + 19² = 2470 -1140

⇒ 1² + 3² + 5² + ... + 19² = 1330

∴ The value of 1² + 3² + 5² + ... + 19² is 1330.

Formula used

Sum of first n odd natural numbers = n²

Calculation

Count of Odd numbers from 1 to 29 = 15 

Sum of first 15 odd natural numbers = 15² = 225

∴ 225

Given:

The numbers are even and lie between 0 and 51

Formula used;

Numbers of terms in AP (n) = [(l - a)/d + 1]

Sum of AP = (n/2) × (a + l)

Here a is first term, l is last term, d is common difference

Calculation:

The numbers are 2, 4, 6, ..... 50

Total even numbers are [(50 - 2)/2 + 1] = 25

Sum of terms = (25/2) × (2 + 50) = 25 × 26

⇒ Sum of terms = 650

∴ 650 is the correct answer.