Question

How many terms of the series 1+2+4+8+........Should be taken so that their sum is 1023?

Answer 1

Step-by-step explanation:

Your mathematical toolkit should include the identity

a^0 + a^1 +…+ a^(n-1) = (1 - a^n) / (1 - a) = (a^n - 1) / (a - 1).

So if a = 2 and n = 10, we have

1+2+4+8+16+32+64+128+256+512 = (2^10 - 1) / (2 - 1) = (1024 - 1) / 1 = 1023.

**Proof**:

Let S(n) = a^0 + a^1 +…+ a^(n-1)

Clearly we can get to S(n+1) by adding the next term, a^n

S(n+1) = (a^0 + a^1 +…+ a^(n-1)) + a^n

= S(n) + a^n.

We can also get it by adding 1 = a^0 and multiplying the remaining terms by a, giving

S(n+1) = a^0 + a * (a^0 + a^1 +…+ a^(n-1))

= 1 + a * S(n).

Equating the two expressions for S(n+1) we get

S(n) + a^n = 1 + a * S(n)

Bringing the terms containing S(n) to one side, we get

S(n) - a * S(n) = 1 - a^n

which is

S(n) * (1 - a) = (1 - a^n)

hence with a not being 1

S(n) = (1 - a^n) / (1 - a)

alternatively

a * S(n) - S(n) = a^n - 1

which is

S(n) * (a - 1) = (a^n - 1)

hence with a not being 1

S(n) = (a^n - 1) / (a - 1)