Question

find the sum of n terms of the series is 1*2*3*+2*3*4+3*4*5+.....

Answer 1

to n th term is n.(n+1).(n+2)

$t_n$ = n³ + 3n² + 2n

so $S_n$ for n = $\frac{n(n+1)}{2}$

$S_n$ for n² = $\frac{n(n+1)(2n+1)}{6}$

$S_n$ for n³ = ${\frac{n(n+1)}{2}}^2$

so  $t_n$ = n³ + 3n² + 2n

$S_n$ = ${\frac{n(n+1)}{2}}^2$ + 3$\frac{n(n+1)(2n+1)}{6}$ + 2$\frac{n(n+1)}{2}$

= $\frac{n^2(n+1)^2}{4}$ + $\frac{n(n+1)(2n+1)}{2}$ + $n(n+1)$

= $n(n+1)[ \frac{n(n+1)}{4}+\frac{2n+1}{2}+1]$

= $n(n+1)[n^2+5n+6)]$

= n(n+1)(n+2)(n+3)

Answer 2

Step-by-step explanation:

solution :

1,2,3,.................are in A. P, 2,3,4 are in A. P

tn=1+(n-1) tn =2-(n-1)

= 1+n-1. = 2+n-1

= n. = n+1

3,4,5.......... are in A.P

tn=3+(n-1)=3+n-1=n+2

nth term = n(n+1) (n+2)

n(n+1).(n+2)=£n(n+2n+n-2)

£n(nsquare+3n square +2n square) =£n square +3n square + 2n square

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

= n(n+1)/2.[n square +5n square +6/2]

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

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