63^26 + 10 13^25 + 10 23^24 + 10 2^23^23 + + 10 2^243 is equal to :

Question

$ \frac{6}{3^{26}} + \frac{10 \cdot 1}{3^{25}} + \frac{10 \cdot 2}{3^{24}} + \frac{10 \cdot 2^2}{3^{23}} + \ldots + \frac{10 \cdot 2^{24}}{3} $ is equal to :

Accepted Answer

Let $\, S=\frac{6}{{3}^{26}}+\frac{10\cdot 1}{{3}^{25}}+\frac{10\cdot 2}{{3}^{24}}+\frac{10\cdot {2}^{2}}{{3}^{23}}+\cdots +\frac{10\cdot {2}^{24}}{3}$

Notice that from the second term onwards, the terms follow a pattern. Let ${S}^{'}$ be the sum of these terms:

${S}^{'}=\frac{10\cdot {2}^{0}}{{3}^{25}}+\frac{10\cdot {2}^{1}}{{3}^{24}}+\frac{10\cdot {2}^{2}}{{3}^{23}}+\cdots +\frac{10\cdot {2}^{24}}{{3}^{1}}$

We can factor out $\frac{10}{{3}^{25}}$ :

$\begin{gathered} {S}^{'}=\frac{10}{{3}^{25}}(1+\frac{{2}^{1}}{{3}^{-1}}+\frac{{2}^{2}}{{3}^{-2}}+\cdots +\frac{{2}^{24}}{{3}^{-24}}) \ {S}^{'}=\frac{10}{{3}^{25}}({2}^{0}\cdot {3}^{0}+{2}^{1}\cdot {3}^{1}+{2}^{2}\cdot {3}^{2}+\cdots +{2}^{24}\cdot {3}^{24}) \end{gathered}$

${S}^{'}=\frac{10}{{3}^{25}}\sum _{k=0}^{24}(2\cdot 3{)}^{k}=\frac{10}{{3}^{25}}\sum _{k=0}^{24}{6}^{k}$

Calculate the sum of the GP

The series $\sum _{k=0}^{24}{6}^{k}$ is a geometric progression with first term $a=1$, common ratio $r=6$, and $n=25$ terms.

The sum formula is ${S}_{n}=a\frac{{r}^{n}-1}{r-1}$ :

$\sum _{k=0}^{24}{6}^{k}=\frac{{6}^{25}-1}{6-1}=\frac{{6}^{25}-1}{5}$

Substituting this back into ${S}^{'}$ :

${S}^{'}=\frac{10}{{3}^{25}}(\frac{{6}^{25}-1}{5})=\frac{2}{{3}^{25}}({6}^{25}-1)=\frac{2\cdot {6}^{25}}{{3}^{25}}-\frac{2}{{3}^{25}}$

Since ${6}^{25}=(2\cdot 3{)}^{25}={2}^{25}\cdot {3}^{25}$ :

${S}^{'}=\frac{2\cdot {2}^{25}\cdot {3}^{25}}{{3}^{25}}-\frac{2}{{3}^{25}}=2\cdot {2}^{25}-\frac{2}{{3}^{25}}={2}^{26}-\frac{2}{{3}^{25}}$

Find the total sum $S$ :

Now, add the first term $\frac{6}{{3}^{26}}$ back to ${S}^{'}$ :

$S=\frac{6}{{3}^{26}}+{2}^{26}-\frac{2}{{3}^{25}}$

[$\frac{6}{{3}^{26}}=\frac{2\cdot 3}{{3}^{26}}=\frac{2}{{3}^{25}}$]

$\Rightarrow$ $S=\frac{2}{{3}^{25}}+{2}^{26}-\frac{2}{{3}^{25}}={2}^{26}$

Solutions · 1