1+2+3+......+ n(n+1)= (n(n+1)(n+2)) / 3
who know how to solve this question,pls show the clear step for me,i am preparing for my test
who know how to solve this question,pls show the clear step for me,i am preparing for my test
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Assume this is true for n=p, then
1+2+3+......+ p(p+1)= (p(p+1)(p+2)) / 3
now put p=p+1 into thr RHS and get
RHS= (p+1)(p+2)(p+3)) / 3
= (p+3)(p+1)(p+2)) / 3, now expand using only the (p+3) factor,
= (p(p+1)(p+2)) / 3 + (p+1)(p+2)
also we know that
(p(p+1)(p+2)) / 3 = 1+2+3+......+ p(p+1), therefore
RHS= 1+2+3+......+ p(p+1) + (p+1)(p+2),
adding the last 2 terms we get
p(p+1) + (p+1)(p+2)= (p+1)(p+2)
therefore
RHS= 1+2+3+......+ p(p+1) + (p+1)(p+2)
which proves the equation
1+2+3+......+ p(p+1)= (p(p+1)(p+2)) / 3
now put p=p+1 into thr RHS and get
RHS= (p+1)(p+2)(p+3)) / 3
= (p+3)(p+1)(p+2)) / 3, now expand using only the (p+3) factor,
= (p(p+1)(p+2)) / 3 + (p+1)(p+2)
also we know that
(p(p+1)(p+2)) / 3 = 1+2+3+......+ p(p+1), therefore
RHS= 1+2+3+......+ p(p+1) + (p+1)(p+2),
adding the last 2 terms we get
p(p+1) + (p+1)(p+2)= (p+1)(p+2)
therefore
RHS= 1+2+3+......+ p(p+1) + (p+1)(p+2)
which proves the equation