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\text{If } T_r = \sqrt{r} \sqrt{r+1} \left( \frac{4r + 5}{(r+2) + \sqrt{r^2 - 1}} \right), \text{ then find the value of } \frac{1}{\sqrt{68}} \sum_{r=1}^{16} T_r

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Published July 1, 2025
Mathematics
Algebra
Sequences & Series
Telescoping Series
Manipulation & Rationalisation

Detailed Explanation

1. Rationalising the ugly fraction\nTrT_r is given as\nTr=r r+1  (4r+5(r+2)+r2−1)T_r = \sqrt{r}\,\sqrt{r+1}\;\Bigg(\frac{4r+5}{(r+2)+\sqrt{r^2-1}}\Bigg)\nIf you multiply top & bottom of the bracket by the conjugate ((r+2) - \sqrt{r^2-1}), the denominator becomes a (a^2 - b^2) form:\n(r+2)2  −  (r2−1)2  =  r2+4r+4  −  (r2−1)  =  4r+5.(r+2)^2 \; - \;(\sqrt{r^2-1})^2\;=\;r^2 +4r+4 \; - \;(r^2-1)\;=\;4r+5.\nBecause (4r+5) also sits in the numerator, they cancel nicely, giving\nTr=r(r+1)[(r+2)−r2−1].T_r = \sqrt{r(r+1)}\Big[(r+2) - \sqrt{r^2-1}\Big].\n\n### 2. Breaking the square-root product\nNotice\nr(r+1) r2−1  =  r(r+1)(r2−1)  =  (r+1)r(r−1).\sqrt{r(r+1)}\,\sqrt{r^2-1}\;=\;\sqrt{r(r+1)(r^2-1)}\;=\;(r+1)\sqrt{r(r-1)}.\nPlugging that back:\nT_r = \big(r+2\big)\sqrt{r(r+1)}\; -\;\big(r+1\big)\sqrt{r(r-1)}.$\n\n### 3. Spotting the telescope\nDefine an auxiliary term\nA_r ;=;\big(r+2\big)\sqrt{r(r+1)}.\nThen\n\[A_r - A_{r-1}\] equals exactly $T_r$ because \n$A_{r-1} = (r+1)\sqrt{(r-1)r}$, which is the second part of $T_r$ with a minus sign.\nHence\nT_r = A_r - A_{r-1}.\n\n### 4. Summing quickly\nWhen you add T_1 + T_2 + \dots + T_{16},everythingcancelsexcept, everything cancels except A_{16}andandA_0.\nBecause.\nBecause A_0 = 2\sqrt{0\cdot1}=0, the total becomes\n$$\sum_{r=1}^{16} T_r = A_{16} = (16+2)\sqrt{16\cdot17} = 18\times4\sqrt{17}=72\sqrt{17}.$$\n\n### 5. Final division\n\sqrt{68}=\sqrt{4\cdot17}=2\sqrt{17}$, so\n168∑r=116Tr=7217217=36.\frac{1}{\sqrt{68}}\sum_{r=1}^{16}T_r = \frac{72\sqrt{17}}{2\sqrt{17}} = 36.\n\nKey concepts used:\n* Conjugate multiplication (rationalisation)\n* Telescoping series\n* Basic surd manipulation

Simple Explanation (ELI5)

What is the question?\nWe have a big looking term TrT_r and we have to add it up from r=1r = 1 to 1616. Then we divide the sum by 68\sqrt{68}.\n\n### How to think about it (like building blocks)\n1. Imagine TrT_r is a Lego made of two smaller Legos.\n2. If we can break it into (something for r) minus the same something for (r!!−!1), then when we add all the TrT_r’s, most Legos cancel out like a sliding set of dominos.\n3. That cancellation trick is called telescoping – only the first and last blocks survive.\n4. Once the sum collapses, the left-over bit is easy to divide by 68\sqrt{68}.\n\nSo the whole problem is really about spotting (or creating) that subtraction pair inside TrT_r.

Step-by-Step Solution

Step-by-step Solution\n\n1. Rationalise the fraction inside TrT_r\n Tr=r r+1(4r+5(r+2)+r2−1)T_r = \sqrt{r}\,\sqrt{r+1}\left(\frac{4r+5}{(r+2)+\sqrt{r^2-1}}\right)\n Multiply numerator & denominator by (r+2)−r2−1(r+2)-\sqrt{r^2-1}:\n Tr=r(r+1) (4r+5)[(r+2)−r2−1](r+2)2−(r2−1)2.T_r = \sqrt{r(r+1)}\,\frac{(4r+5)\big[(r+2)-\sqrt{r^2-1}\big]}{(r+2)^2-(\sqrt{r^2-1})^2}.\n Denominator ((r+2)^2-(r^2-1)=4r+5) cancels the numerator (4r+5), leaving\n Tr=r(r+1)[(r+2)−r2−1].T_r = \sqrt{r(r+1)}\Big[(r+2) - \sqrt{r^2-1}\Big].\n\n2. Rewrite the square-root product\n r(r+1)r2−1=r(r+1)(r2−1)=(r+1)r(r−1).\sqrt{r(r+1)}\sqrt{r^2-1}=\sqrt{r(r+1)(r^2-1)}=(r+1)\sqrt{r(r-1)}.\n Therefore,\n T_r = (r+2)\sqrt{r(r+1)} - (r+1)\sqrt{r(r-1)}.$\n\n3. **Create a telescoping expression**\n Define\n A_r = (r+2)\sqrt{r(r+1)}.\nThen\n\n Then\n A_{r-1} = (r+1)\sqrt{(r-1)r},\nand\n\n and\n T_r = A_r - A_{r-1}.\n\n4. **Add from $r=1$ to $16$**\n \[\sum_{r=1}^{16} T_r = \sum_{r=1}^{16} (A_r - A_{r-1}) = (A_1-A_0)+(A_2-A_1)+\dots+(A_{16}-A_{15}) = A_{16} - A_0.\]\n But $A_0 = 2\sqrt{0\cdot1}=0$, hence\n \sum_{r=1}^{16} T_r = A_{16} = (16+2)\sqrt{16\cdot17}=18\times4\sqrt{17}=72\sqrt{17}.\n\n5. **Divide by $\sqrt{68}$**\n \sqrt{68}=\sqrt{4\cdot17}=2\sqrt{17},\n\n \frac{1}{\sqrt{68}}\sum_{r=1}^{16} T_r = \frac{72\sqrt{17}}{2\sqrt{17}}=36.$$\n\n[\boxed{36}]

Examples

Example 1

Cancelling out arrays of resistors in physics circuits—series resistances add like a telescoping sum.

Example 2

Financial EMI tables where only first and last instalment amounts matter after netting off intermediary charges.

Example 3

Computer algorithms that simplify cumulative differences, e.g., prefix sums subtracting adjacent terms.

Visual Representation

References

  • [1]I.A. Maron – Problems in Calculus of One Variable (Sequence manipulations chapter)
  • [2]Arihant Skills in Mathematics: Algebra for JEE Main & Advanced – Telescoping sums section
  • [3]MIT OpenCourseWare – Algebraic Manipulation Techniques video series
  • [4]Brilliant.org articles on Telescoping Series

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