Chapter 12.3, Problem 70E

To determine

Calculate the sum of the first ten terms of the sequence.

Answer to Problem 70E

The sum of the first ten terms is $\frac{a(1-a^{100})}{1-a}+55b$ .

Explanation of Solution

Given:

It is given in the concept that the terms ar $(a+b),(a^2+2b),(a^3+3b),(a^4+4b),\mathrm{............}(a^{100}+10b)$ .

Concept Used:

In this, use the concept of geometric series and arithmetic series by understanding the pattern of the terms and the formula used is $S_n=\frac{n}{2}[2a+(n-1)d]and{S}_n=\frac{a(1-r^n)}{1-r}$ .

Calculation:

Here, the terms are $(a+b),(a^2+2b),(a^3+3b),(a^4+4b),\mathrm{............}(a^{100}+10b)$ .

Now, break down the sum of the first ten terms

  $(a+b),(a^2+2b),(a^3+3b),(a^4+4b),\mathrm{............}$ to the sum of $a+a^2+\mathrm{.........}+a^{10}$ and $b+2b+3b+\mathrm{..........}+10b$ .

The former is the partial sum of an arithmetic series and the latter is the partial sum of an arithmetic series.

  $a+a^2+\mathrm{.........}+a^{10}=\frac{a(1-a^{100})}{1-a}$

  $b+2b+3b+\mathrm{..........}+10b=\frac{10}{2}[2b+(10-1)b]=55b$

The sum is then, $\frac{a(1-a^{100})}{1-a}+55b$ .

Conclusion:

The sum is $\frac{a(1-a^{100})}{1-a}+55b$ .