Question

the sum of all different products formed by first four natural numbers taken two at a time​

Answer 1

Answer:

Step-by-step explanation:

Let the required sum be S.

(1+2+3+...+n)2=12+22+32+42+...+n2+2S.  

But  (1+2+3+...+n)2=[n(n+1)2]2.  

12+22+32+42+...+n2=n(n+1)(2n+1)6.  

So

S=n(n+1)(n−1)(3n+2)24.  

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Ashish Singh, worked at Infosys

Answered September 12, 2015 · Author has 394 answers and 1.1M answer views

ok so you mean to find for even "n"

1.2 + 3.4 +...+ (2n-1).(2n)

the general term is

T(n) =  (2n−1)∗(2n)  

=  4n2−2n  

now summing that up is not very difficult

so sum of n terms of this series gives

n(n+1)(4n−1)3  

now as you want to get sum for first n natural numbers , just replace n by n/2 as first n natural numbers are covered in first n/2 terms

so your required sum is

n(n+2)(2n−1)12