Question

1+4+7+10+13+.......+94+97+100=?​

Answer 1

1+4+7+10+13+17+21+25+29+33+47+51+55+59+63+67+71+75+79+83+87+91+

А дальше сам(-а)

Answer 2

The given sequence is an arithmetic progression with a common difference of 3. We can find the sum of an arithmetic progression using the following formula:

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

where: Sn is the sum of the arithmetic progression n is the number of terms in the progression a is the first term of the progression d is the common difference

In this case, the first term (a) is 1, the common difference (d) is 3, and we need to find the sum up to the last term, which is 100.

Using the formula:

Sn = (n/2)(2a + (n-1)d) = (n/2)(2(1) + (n-1)(3)) = (n/2)(2 + 3n - 3) = (n/2)(3n - 1)

We need to find the value of n, which represents the number of terms in the sequence. We can use the formula for the nth term of an arithmetic progression to find n:

an = a + (n-1)d

In this case, we know that the last term is 100. Substituting the values into the formula:

100 = 1 + (n-1)3 99 = 3n - 3 102 = 3n n = 34

Now that we know n, we can substitute it back into the formula for Sn:

Sn = (n/2)(3n - 1) = (34/2)(3(34) - 1) = 17(102 - 1) = 17(101) = 1717

Therefore, the sum of the given sequence is 1717.