Question

1.2+2.3+3.4+...+n(n+1)=

Answer 1

let s=1.2+2.3+3.4+....+n(n+1)

s=∑n(n+1)   (∑=summation)

s=∑(n²+n)=∑n²+∑n        (I have just split the summation expression)

we know ∑n²=sum of squares of first n natural numbers=n(2n+1)(n+1)/6

and ∑n=sum of first n natural numbers=$(\frac{n(n+1)}{2}</p><p>∴ s=[tex]\frac{ n(2n+1)(n+1)}{6}   + \frac{ n(n+1)}{2}$

    = ($(\frac{n(n+1)}{2})(\frac{2n+1}{3} +1)$

   s=$\frac{n(n+1)(n+2)}{3}$

Answer 2

1.2 + 2.3 + 3.4 +………….. to n terms

The nth term = n(n+1) = n² + n

n n²

1 1

2 4

3 9

n n²

the sum = ∑ n² + ∑n = n(n+1)(2n+1)/6 + n(n+1)/2 = n(n+1)/2 . [(2n+1)/3 + 1]

= n(n+1)/2 . (2n+4)/3 = n(n+1)/2 . 2(n+2)/3

= n(n+1) (n+2)/3

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