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Grade: 12
         the sum of n terms of series:

Q.1>1^2+2*2^2+3^2+2*4^2+5^2+2*6^2+........


Q.2> The number of proper divisiors of 2^p*6^q*15^r
7 years ago

Answers : (1)

Pramod J AskiitiansExpert-IIT-B
36 Points
							

Q.1>1^2+2*2^2+3^2+2*4^2+5^2+2*6^2+........

sum can be divided into 2 : 1^2 + 3^2 + 5^2 +..... + 2*[ 2^2 + 4^2 +......]

=> 1^2 + 2^2 + 3^2 +...... + n^2 +[ 2^2 + 4^2 +.....]

if n is even,

=> n*(n+1)*(2n+1)/6 + 4*[ (n/2)*(n/2 +1)*(n+1)/6]

if n is odd,

=> n*(n+1)*(2n+1)/6 + 4*[ (n-1/2)*(n-1/2 +1)*(n)/6]

 

Q.2> The number of proper divisiors of 2^p*6^q*15^r

the above number can be written as 2^p*(2*3)^q*(3*5)^r => 2^(p+q)*3^(q+r)*5^r

so, the number of proper divisors are (p+q)*(q+r)*r



7 years ago
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