Question

2+4+6+8.....+128. Помогите решить

Answer 1

2+4+6+8+(10+12+14+16+17+18+20+22+24+26+28+30+32+34+36+38+40+42+44+46+48+50+52+54+56+58+60+62+64+66+68+70+72+74+76+78+80+82+84+86+88+90+92+94+96+98+100+102+104+106+108+110+112+114+116+118+120+122+124+126)+128 = 4160

Answer 2

Solution for the Series 2, 4, 6, 8..... 128

To find the sum of the series 2, 4, 6, 8..... 128, we can use the formula for the sum of an arithmetic series:

Sum of an arithmetic series = (n/2) * (first term + last term)

Where: - n = number of terms - first term = 2 - last term = 128

Calculating the Number of Terms (n)

To find the number of terms in the series, we can use the formula: nth term = first term + (n-1) * common difference

In this case, the common difference is 2, and we need to find the value of n when the nth term is 128.

Using the formula: 128 = 2 + (n-1) * 2 Solving for n: 128 - 2 = 2n - 2 126 = 2n - 2 2n = 126 + 2 2n = 128 n = 128 / 2 n = 64

So, the number of terms (n) in the series is 64.

Calculating the Sum

Now, we can calculate the sum of the series using the formula: Sum of an arithmetic series = (n/2) * (first term + last term)

Substituting the values: Sum = (64/2) * (2 + 128) Sum = 32 * 130 Sum = 4160

Therefore, the sum of the series 2, 4, 6, 8..... 128 is 4160.

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