Skip to content
Exercise 5.4 · Q10

Q.Find the value of ∑n=1∞12n−1(19n−1+192n−1)\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{2^{n-1}}\left(\dfrac{1}{9^{n-1}}+\dfrac{1}{9^{2n-1}}\right).

Tamil Nadu DgeTextbookSubjectiveImportance★★★★★est
51% · 48/95 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Split the series into two separate geometric series (one from each term inside the bracket), sum each with the infinite-GP formula, and add.

Step 1. Split the sum.

∑n=1∞12n−1(19n−1+192n−1)=∑n=1∞12n−19n−1+∑n=1∞12n−192n−1.\sum_{n=1}^\infty\frac1{2^{n-1}}\left(\frac1{9^{n-1}}+\frac1{9^{2n-1}}\right) = \sum_{n=1}^\infty\frac1{2^{n-1}9^{n-1}} + \sum_{n=1}^\infty\frac1{2^{n-1}9^{2n-1}}.

Step 2. Sum the first GP. 12n−19n−1=(118)n−1\dfrac1{2^{n-1}9^{n-1}}=\left(\dfrac1{18}\right)^{n-1}, a GP with ratio 118\dfrac1{18}:

∑n=1∞(118)n−1=11−118=1817.\sum_{n=1}^\infty\left(\frac1{18}\right)^{n-1} = \frac1{1-\frac1{18}} = \frac{18}{17}.

Step 3. Sum the second GP. Write 92n−1=81n99^{2n-1}=\dfrac{81^n}9, so 12n−192n−1=92n−181n=19(12⋅81)n−1=19(1162)n−1\dfrac1{2^{n-1}9^{2n-1}}=\dfrac9{2^{n-1}81^n}=\dfrac19\left(\dfrac1{2\cdot81}\right)^{n-1}=\dfrac19\left(\dfrac1{162}\right)^{n-1}, a GP with ratio 1162\dfrac1{162}: …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.