Question

T1+T2+....+Tn=n(n+1)(n+2)/6 where tn is the nth triangular number as discussed in the first class.

Answer 1

Given :  Tn = nth triangular number  

To find : show that T1+T2+....+Tn=n(n+1)(n+2)/6

Solution:

nth triangular number   Tn  = ∑n  = n(n+1)/2

T1+T2+....+Tn

=  ∑ Tn

 = ∑ n(n+1)/2

= (1/2)∑(n² + n)

= (1/2)∑n² + (1/2)∑n

= (1/2)n(n+1)(2n+1)/6 + (1/2)n(n+1)/2

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

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

=  (1/12)n(n+1) (2n+4)

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

= (1/6)n(n+1) (n+2)

= n(n+1) (n+2)/6

QED

T1+T2+....+Tn  = n(n+1) (n+2)/6

Hence proved

Learn More:

श्रेणी के n पदों का योग ज्ञात कीजिए ...

https://brainly.in/question/15649136

प्रश्न 1 से 7 तक प्रत्येक श्रेणी के n ...

https://brainly.in/question/9240448