prove that 1+2+ ....... +n = n(n+1)/2?

Answer 1

for all natural numbers n; call this statement P(n). (This is a special case of Faulhaber's formula.) This is a simple formula for the sum of the positive natural numbers less than or equal to number n. The proof that the statement is true for all natural numbers n proceeds as follows.

Check if it is true for n = 0. Notice that the empty sum is 0. And 0(0 + 1) / 2 = 0. So the statement is true for n = 0. Thus we have that P(0) holds.

Now we have to show that if the statement holds when n = m, then it also holds when n = m + 1. This can be done as follows.

Assume the statement is true for n = m, i.e.,


Adding m + 1 (which is clearly the left-hand side's next term) to both sides gives


By algebraic manipulation we have for the right-hand side


Thus we have


This is just P(m + 1). This proof is conditional: we made the assumption that P(m) is true, and from that assumption we derived P(m + 1). Symbolically, we have shown that


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

Answer 2

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

Answer 3

1+2=?

by using the formula

2(2+1)/2
therefore 2+1=3.

Answer 4

For the base case, you will probably want to start with n = 3 (i.e., when you have at least 3 elements in your set to begin with). However, note that the formula agrees with the description when n = 0, 1, 2, as surely none of these sets have a 3-element subset. For the induction, let S be a given n+1-element set, and focus on an element a in S. Then any 3-element subset of S either contains a, and thus the remaining elements form a 2-element subset of S \ {a}, or it does not contain a, and thus the 3-element subset is in fact a 3-element subset of S \ {a}.

Answer 5

If you know the identity and simply mean to verify it, a proof by mathematical induction will suffice. Simply assume true for some n=p and prove that the identity holds true for n=p+1, now verify with n=1 and you have proven the identity for all positive integers.
On the other hand, if you wish to derive it, the most straight forward method is to consider twice the sum. i.e.
S = 1+2+...+n = n+...+2+1
2S = n(n+1)
S = (n/2).(n+1)

Hope this helps!

Answer 6

1 2 3 4 5 6 7 8 n
n n-1 n-2 n-3 n-4 n-5 n-6 n-7 1

the sum of each digit in line one with bottom digit in line 2
are equal. n+1=2+(n-1)=3+(n-2)=4+(n-3)=...=n+1
so the sum of these 2 lines will be n(n+1)
so the sum of one line (means from 1 to n) will be n(n+1)/2

Answer 7

Let n be 1
Then the answer is obvious(i.e.1(1+1)/2=1)
Let n be 2
Then the answer is 2(2+1)/2=3
Suppose the answer is true for n
i.e.n(n+1)/2 is true
We are going to prove for n+1
i.e.
n+1(n+2)/2 is true----------(*)
proof:
1+2+3+...+n=n(n+1)/2
2(1+2+3+...+n)=n(n+1)
2(1+2+3+...+n)/n+1=n------(1)
But2(1+2+3+...+n+n+1)/n+2=n+1--------from(*)--------(2)
from 1 and 2
if it is for n terms r.h.s is n and if it is for n+1 terms r.h.s. is n+1
Therefore our result is proved
i.e.1+2+3+...+n=n(n+1)/2 is true for all n in natural numbers

Answer 8

1 + 2+ 3+ 4+.......+100 ---Equation1
99+98+97+96+.......+ 1 ---Equation2
Add1&2Every where you get101for 100 timesBecause i added it twice divide 101*100by 2
suppose we have n numbers then it is (n+1)*n divided by 2

Answer 9

Let the sum of 'n'th term of the series = Sn

Sn = 1 + 2 + 3 + ... + (n - 1) + n .................(1)

Rewrite (1) backwards,
Sn = n + (n - 1) + (n - 2) + ... + 2 + 1 ................(2)

Adding up (1) & (2);
2Sn = (1+n) + (1+n) + (1+n) ...

Since you have n terms,
2Sn = n (1+n)
Therefore, Sn = n(n+1)/2

Hence, 1+2+...+n=n(n+1)/2

Answer 10

Sn = 1+2+3+...+n......(1)
Sn = n+(n-1)+...+1......(2)

(1)+(2): 2Sn = n(n+1)

Therefore, Sn = n(n+1)/2