126. The value of 5² + 6² + ..... + 10² + 20² is
     

1. 1. 755
      

   
   
   

   2. 760
      

   
   
   

   3. 765
      

   
   
   

   4. 770

   
   
   

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1. View Hint View Answer Discuss in Forum

   We know that ,
   

   1² + 2² + 3² + ..... + n² =	n(n + 1)(2n + 1)
   	6

   
   

   Correct Option: A

   We know that ,
   

   1² + 2² + 3² + ..... + n² =	n(n + 1)(2n + 1)
   	6

   
   

   Required answer =	10 × 11 ×21	+ 20² -	4 × 5 ×9
   	6		6

   
   Hence , Required answer = 385 + 400 – 30 = 755

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127. 1² – 2² + 3² – 4² + .... – 10² is equal to
     

1. 1. 45
      

   
   
   

   2. –45
      

   
   
   

   3. –54
      

   
   
   

   4. –55

   
   
   

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1. View Hint View Answer Discuss in Forum

   Suppose S = 1² – 2² + 3² – 4² + ... – 10²
   S = (1² + 3² + 5² + 7² + 9²) – (2² + 4² + 6² + 8² + 10²)
   We know that ,
   

   Sum of squares of first n odd natural numbers =	n(4n² - 1)
   	3

   
   

   Sum of squares of first n even natural numbers =	2	n(n + 1)(2n + 1)
   	3

   
   

   Hence, S =	5(4 × 5 × 5 - 1)	-	2	× 5(5 + 1) (2 × 5 + 1)
   	3		3

   
   

   Correct Option: D

   Suppose S = 1² – 2² + 3² – 4² + ... – 10²
   S = (1² + 3² + 5² + 7² + 9²) – (2² + 4² + 6² + 8² + 10²)
   We know that ,
   

   Sum of squares of first n odd natural numbers =	n(4n² - 1)
   	3

   
   

   Sum of squares of first n even natural numbers =	2	n(n + 1)(2n + 1)
   	3

   
   

   Hence, S =	5(4 × 5 × 5 - 1)	-	2	× 5(5 + 1) (2 × 5 + 1)
   	3		3

   
   

   S =	5 × 99	=	2 × 30 × 11
   	3		3

   
   Therefore , S = 165 – 220 = – 55

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128. (1² + 2² + 3² + .......... + 10²) is equal to
     

1. 1. 380
      

   
   
   

   2. 385
      

   
   
   

   3. 390
      

   
   
   

   4. 392

   
   
   

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1. View Hint View Answer Discuss in Forum

   Using the given formula ,
   

   1² + 2² + 3² + ... + n² =	n(n + 1)(2n + 1)
   	6

   
   Here , n = 10
   

   Correct Option: B

   Using the given formula ,
   

   1² + 2² + 3² + ... + n² =	n(n + 1)(2n + 1)
   	6

   
   Here , n = 10
   

   ∴ 1² + 2² + 3² + .... + 10² =	10(10 + 1)(20 + 1)	= 385
   	6

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129. (5² + 6² + 7² + ... + 10²) is equal to
     

1. 1. 330
      

   
   
   

   2. 345
      

   
   
   

   3. 355
      

   
   
   

   4. 360

   
   
   

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1. View Hint View Answer Discuss in Forum

   Using the given formula ,
   

   1² + 2² + 3² + 4² + ... + n² =	n(n + 1)(2n + 1)
   	6

   
   ∴ 5² + 6² + .... 10² = (1² + 2² + ... + 10²) – (1² + 2² + 3² + 4²)
   Here , n = 10 and n = 4
   

   Correct Option: C

   Using the given formula ,
   

   1² + 2² + 3² + 4² + ... + n² =	n(n + 1)(2n + 1)
   	6

   
   ∴ 5² + 6² + .... 10² = (1² + 2² + ... + 10²) – (1² + 2² + 3² + 4²)
   Here , n = 10 and n = 4
   

   Required answer =	10 × 11 × 21	-	4 × 5 × 9
   	6		6

   
   Required answer = 385 – 30 = 355

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130. [2² + 3² + 4² + 5² + 6² +7² +8² + 9² +10² ] is equal to
     

1. 1. 385
      

   
   
   

   2. 2916
      

   
   
   

   3. 540
      

   
   
   

   4. 384

   
   
   

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1. View Hint View Answer Discuss in Forum

   We know that ,
   

   1² + 2² + 3² + .... + n² =	n(n + 1)(2n + 1)
   	6

   
   ∴ 2² + 3² + 4² + ..... + 10² = (1² + 2² + 3² + ..... + 10²) – 1
   Here , n = 10
   

   Required answer =	10(10 + 1)(2 × 10 + 1)	- 1
   	6

   
   

   Correct Option: D

   We know that ,
   

   1² + 2² + 3² + .... + n² =	n(n + 1)(2n + 1)
   	6

   
   ∴ 2² + 3² + 4² + ..... + 10² = (1² + 2² + 3² + ..... + 10²) – 1
   Here , n = 10
   

   Required answer =	10(10 + 1)(2 × 10 + 1)	- 1
   	6

   
   

   Required answer =	10 × 11 × 21	- 1 = 385 - 1 = 384
   	6

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