[" The sum of the first "n" terms of the...

[" The sum of the first "n" terms of the series "],[1^(2)+22^(2)+3^(2)+24^(2)+5^(2)+2*6^(2)+..." is "(n(n+1)^(2))/(2)," when "],[n" is even.When "n" is odd,the sum is ".....]

Similar Questions

Explore conceptually related problems

The sum of first n terms of the series 1^(2) + 2.2^(2) +3^(2) + 2. 4^(2) + 5^(2) + 2. 6^(2) + "........." is ( n ( n + 1)^(2))/( 2) when n is even. When, n is odd, the sum is :

The sum of n terms of the series 1^(2)+2.2^(2)+3^(2)+2.4^(2)+5^(2)+2.6^(2)+... is (n(n+1)^(2))/(2) when n is even.when n is odd the sum is

The sum of first n terms of the series 1_(2) + 2.2^(2) + 3^(2) + 2.4^(2) + 5^(2) + 2.6^(2) + ...... is (n(n + 1)^(2))/4 when n is even. When n odd the sum is

The sum of first n terms of the series 1^(2) + 2.2^(2) + 3^(2) + 2.4^(2) + 5^(2) + 2.6^(2) + ….is (n(n+1)^(2))/(2) when n is odd the sum is

The sum of n terms of the series 1^(2) - 2^(2) + 3^(2)- 4^(2) + 5^(2) -6^(2) + ….is

The sum of n terms of the series 1^(2)-2^(2)+3^(2)-4^(2)+5^(2)-6^(2)+... is

The sum of first n terms of the series 1^(2)+ 2 . 2^(2) + 3^(2) + 2. 4 ^(2) + 5^(2) +… is n/2 (n+1) ^(2) when n is even , When n is odd then the sum will be-

[" The sum of the first "n" terms of the series "],[1^(2)+2*2^(2)+3^(2)+2*4^(2)+5^(2)+2*6^(2)+..." is "(n(n+1)^(2))/(2)," when "],[n" is even.When "n" is odd,the sum is ".....]

Similar Questions

Recommended Questions

[" The sum of the first "n" terms of the series "],[1^(2)+22^(2)+3^(2)+24^(2)+5^(2)+2*6^(2)+..." is "(n(n+1)^(2))/(2)," when "],[n" is even.When "n" is odd,the sum is ".....]