Question

find the sum of 1+ 2 +3+4.....to n terms.​

Answer 1

$S_{n} = 3 +8 + 13 + 18 ....+ n terms\\therefore ,\\ a = 3\\ d = 8 - 3 = 5\\, \\\\a_{n} = a + (n-1)d\\ = 3 +( n-1 )5\\ = 5n-2\\therefore,\\ s_{n} = [\frac{a + a_{n} }{2} n\\ = \frac{n}{2} [ 3 + 5n -2]\\ = \frac{n}{2} (5n+1)$

Answer 2

Answer:

n² + 4n is the answer

Step-by-step explanation:

So when we add this sequence we will get an AP

3,5,7,9.........

So now we can Calculate it by using Sn Formula

$Sn = n/2 + ( 2a + ( n-1 ) d )$

$Sn = n/2 + ( 2(3) + (n-1)2)\\Sn = n/2 + ( 6 + 2n - 2 )\\= n/2 + ( 4 + 2n ) \\n{2} + 4n$

Hope it helped you

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