Question

mathematical induction 1+2+3+....+n=1/2(n+1) prove that (answer assignment)

Answer 1

friend it is not 1+2+3+.......+n=1/2(n+1) it is 1+2+3+........+n=n/2(n+1).

1+2+3+........+n not equals to 1/2(n+1)
proof:
let n=7
then,
LHS=1+2+3+....+7=28..........................1
RHS=1/2(7+1)=8/2=4............................2
By equation 1 and 2 we get,
LHS not equals to RHS
i.e 1+2+3+....+n not equals to 1/2(n+1)

Now: 1+2+3+......n=n/2(n+1)
Proof:
let n=7
RHS:n/2(n+1)=7/2(8)=28.................3
by equation 1 and 3 we get,
1+2+3+.........n=n/2(n+1)
now let n=18
LHS=1+2+..........+18=171
RHS=n/2(n+1)=18/2(18+1)=9(19)=171
By above we get,
LHS=RHS
let us assume,
1+2+3+....n=n/2(n+1).........4 is true for n=k then,, we get,
1+2+3+....+k=k/2(k+1).........5
if this is true, then above statement will also be true for (k+1), then we get,
LHS=1+2+3......k+k+1
=k/2(k+1)+k+1 (by eq. 5)
=[k(k+1)+2(k+1)]/2
=[(k+1)(k+2)]/2
RHS=(k+1)[k+1+1]/2 (by putting n=k+1 in eq. 4)
RHS=[(k+1)(k+2)]/2
by above we get
LHS =RHS
Hence,
1+2+3+.....+n=n/2(n+1) is true for any value of n.


PROGRESSION METHOD:
1+2+3+.........+n
above series forms an arithmetic progression.
here first term=a=1, common difference=d=1
last term=n
let nth term of this AP be n
then by the formula anth term=a+(m-1)d, where m is no of terms,, we get,
no. of terms=m=n
above AP shows sum of observations
therefore, 1+2+3+....+n=n/2(2×1+(n-1)1)
=n/2(2+n-1)
=n/2(n+1)
Here formula used:n/2(2a+(n-1)d)