Find sum to n terms of
the following series:

1

^{2 }+ (1^{2}+2^{2}) + (1^{2}+2^{2}+3^{2}) +……….
(1) n(n+1)(n+2)

^{2}/12
(2) n

^{2}(n+1)(n+2)/12
(3) n(n+1)

^{2}(n+2)/12
(4) n(n+1)(2n+1)/6

(5) n

^{2}(n+1)^{2}/4
Answer
follows here:

**Solution:**
The formulae to be used here
are:

Sum of ‘n’ consecutive
integers starting from 1 = 1+2+3+…..+n = ∑n = n(n+1)/2

Sum of squares of ‘n’
consecutive integers starting from 1

= 1

^{2}+2^{2}+3^{2}+…..+n^{2}= ∑n^{2}= n(n+1)(2n+1)/6
Sum of cubes of ‘n’
consecutive integers starting from 1

= 1

^{3}+2^{3}+3^{3}+…..+n^{3}= ∑n^{3}= n^{2}(n+1)^{2}/4
Given series is:

1

^{2 }+ (1^{2}+2^{2}) + (1^{2}+2^{2}+3^{2}) +……….
n

^{th}term of the series = t_{n}= 1^{2}+2^{2}+3^{2}+…..+n^{2}= ∑n^{2}= n(n+1)(2n+1)/6
Sum to n terms = ∑t

_{n}= ∑n(n+1)(2n+1)/6
= 1/6 ∑(n

^{2}+n)(2n+1)
= 1/6 ∑(2n

^{3}+ n^{2}+2n^{2}+n)
= 1/6 ∑(2n

^{3}+3n^{2}+n)
= 1/3 ∑n

^{3 }+ 1/2 ∑n^{2 }+ 1/6 ∑n
= 1/3 (n

^{2}(n+1)^{2}/4) + ½ (n(n+1)(2n+1)/6) + 1/6 (n(n+1)/2)
= n(n+1)/12 {n(n+1)+ (2n+1)
+ 1}

= n(n+1)/12 {n

^{2}+n+2n+2}
= n(n+1)/12 {n(n+1)+2(n+1)}

= n(n+1)/12 {(n+1)(n+2)}

= n(n+1)

^{2}(n+2)/12**Answer (3)**

__Short cut method:__
As “None of these” option
is not given, we can safely check the answer options by substituting value of ‘n’:

For n =1, LHS = 1

^{2}= 1
For RHS answer options can
be checked one by one:

Option(1) ---- 1(1+1)(1+2)

^{2}/12 = 1*2*9/12 ≠ 1 => this option is wrong
Option(2) ---- 1

^{2}(1+1)(1+2)/12 = 1*2*3/12 ≠ 1 => this option is wrong
Option(3) ---- 1(1+1)

^{2}(1+2)/12 = 1*4*3/12 = 1 => this option is correct
Option(4) ---- 1(1+1)(2(1)+1)/6 = 1*2*3/6
= 1 => this option is correct

Option(5) ---- 1

^{2}(1+1)^{2}/4 = 1*4/4 = 1 => this option is correct
Left out options are
(3),(4) and (5).

Now we check for n=2:

LHS = 1

^{2}+ (1^{2}+2^{2})= 1+5 = 6
Option(3) ---- 2(2+1)

^{2}(2+2)/12 = 2*9*4/12 = 6 => this option is correct
Option(4) ---- 2(2+1)(2(2)+1)/6 = 2*3*5/6
≠ 6 => this option is wrong

Option(5) ---- 2

The only left out option is (3)^{2}(2+1)^{2}/4 = 4*9/4 ≠ 6 => this option is wrong
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