Question

3 (3' – 2) 4. i=0

Answer 1

We want to evaluate

$\sum_{i=0}^{i=6} (3^{i} - 2^{i} )$,

There are finite terms in the given expression, thus we can rearrange the terms.

Therefore

.

1=6 =6 Σ(3 – 2) =Σ3 -Σ2 =0 1-0 =0

Notice that both term are a geometric series.

As for any geometric series

$a, ar, ar^{2}, \cdots ar^{n-1}, \,\ r>1,$

we have the sum of series as

$S=a+ ar+ ar^{2}+\cdots +ar^{n-1} =\frac{a( r^{n}-1)}{r-1}, \,\ r>1,$

Thus

,

3. 1.(37 - 1) 3-1 37 -1 2 r = 3, a = 1.

And

$\sum_{i=0}^{i=6} 2^{i} = \frac{1.(2^{7} -1)}{2-1}=\frac{ 2^{7}-1}{1}; r=2, a=1.$

Now

$\sum_{i=0}^{i=6} 3^{i} - \sum_{i=0}^{i=6} 2^{i}= \frac{3^{7}-1}{2}- (2^{7}-1),$

Therefore

$\sum_{i=0}^{i=6} (3^{i} - 2^{i} )=966.$