Using the principle ofmathematical induction prove that`1+1/(1+2)+1/(1+2+3)+1/(1+2+3+4)++1/(1+2+3++n)=(2n)/(n+1)`for all `n in N`

Using the principle ofmathematical induction prove that`1+1/(1+2)+1/(1

1.	Using the principle ofmathematical induction prove that`1+1/(1+2)+1/(1+2+3)+1/(1+2+3+4)++1/(1+2+3++n)=(2n)/(n+1)`for all `n in N`
Answer» `P(n) : 1 + (1)/(1+2) + (1)/(1+2+3) + "…."+(1)/(1+2+3+"…."+n) = (2n)/(n+1)` For `n = 1`, ` L.H.S. = 1` and `R.H.S. = (2 xx 1)/(1+1) = 2/2 = 1` Thus, ` P(1)` is true. Let `P(n)` be true for some `n = k`. i.e, `1+(1)/(1+2)+"……"+(1)/(1+2+3)+"......"+(1)/(1+2+3+"...."k) = (2k)/(k+1)"....."(1)` Now, we have to prove that `P(n)` is true for `n = k + 1`. ie., `1+(1)/(1+2) +"..."+(1)/(1+2+3)+"..."+(1)/(1+2+3+"....."+(k+1))` `= (2(k+1))/(k+2)` Adding `(1)/(1+2+3+"......"+(k+1))` on both sides of `(1)`, we get `1+(1)/(1+2)+"...."+(1)/(1+2+3) +"...."(1)/(1+2+3+"....."+k) + (1)/(1+2+3+"...."+(k+1))` ` = (2k)/(k+1)+(1)/(1+2+3+"....."+(k+1))` `= (2k)/(k+1)+(1)/(((k+1)(k+2))/(2))` [Using `1+2+3+"...."+n = (n(n+1))/(2)`] `= (2k)/((k+1)) + (2)/((k+1)(k+2))` `= (2)/(k+1)((k(k+2)+1)/(k+2))` `= (2(k+1)^(2))/((k+1)(k+2))` `= (2(k+1))/((k+2))` Thus, `P(k+1)` is the whenever `P(k)` is true. Hence by the principle of mathemetical induction, statement `P(n)` is true for all natural numbers.