Question

Prove $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2^n}=1-\frac{1}{2^n}$.

Answer 1

In this we have to prove Left hand side(L.H.S)=Right hand side(R.HS)

Taking L.H.S:-

$\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+................+{\frac{1}{2}}^n$

We know that in Geometric progression(G.P), sum of n terms is given by :-

Sum$=a\times \left(\right(r^n)-1)\frac{1}{r-1}$

Where a is the first term of G.P,

r is the common ratio (i.e second term divided by first term)

n is the number of terms

From L.H.S we have:-

$a=\frac{1}{2}$

$r=\frac{1}{4}\frac{2}{1}$

Putting in formula

$sum=\frac{1}{2}\times \left(\right({\frac{1}{2}}^n)-1)\times \frac{1}{1/2-1}$

On solving we get:-

$L.H.S=1-{\frac{1}{2}}^n$

Which is equal to R.H.S

Hence proved that L.H.S=R.H.S