Question

partial fraction of 1/n(n+1)(n+2)(n+3)

Answer 1

ANSWER

YOU CAN USE Vn METHOD

WHICHNIS USED IN SEQUENCE AND SERIES TO FIND SUM

SEE ATTACHMENT

FOLLOW ME

BYE

Answer 2

Answer:

1/n^2+3n×1/n+2×1/n+1 Or, 1/n×1/n+1×1/n+2×1/n+3

Step-by-step explanation:

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

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

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

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

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

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

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

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

=1/n^2+3n×1/n+2×1/n+1

It can be more short by the following process

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

=1/n×1/n+1×1/n+2×1/n+3