Newton’s Theorem Recurrence Relation Shortcut

 

❓ Question

If $\alpha,\beta$ are roots of:

$$x^2+5\sqrt2\,x+10=0$$

and

$$P_n=\alpha^n-\beta^n$$

then find the value of:

$$\left( \frac{ P_{17}P_{20}+5\sqrt2\,P_{17}P_{19} }{ P_{18}P_{19}+5\sqrt2\,P_{18}^2 } \right)$$

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🖼 Question Image

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✍️ Short Concept

This question is based on:

👉 Newton’s recurrence relation
👉 Quadratic roots sequence
👉 Smart factorisation.

Main trick:

Convert everything into recurrence form.

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🔷 Step 1 — Apply Newton’s Relation 💯

Given quadratic:

$$x^2+5\sqrt2\,x+10=0$$

For sequence:

$$P_n=\alpha^n-\beta^n$$

recurrence relation becomes:

$$P_n+5\sqrt2\,P_{n-1}+10P_{n-2}=0$$

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🔷 Step 2 — Write Required Relations

For $n=20$:

$$P_{20}+5\sqrt2\,P_{19}+10P_{18}=0$$
$$P_{20}+5\sqrt2\,P_{19}=-10P_{18}$$

For $n=19$:

$$P_{19}+5\sqrt2\,P_{18}+10P_{17}=0$$
$$P_{19}+5\sqrt2\,P_{18}=-10P_{17}$$

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🔷 Step 3 — Factorise Expression

Given expression:

$$E= \frac{ P_{17}P_{20}+5\sqrt2\,P_{17}P_{19} }{ P_{18}P_{19}+5\sqrt2\,P_{18}^2 }$$

Take common terms:

$$E= \frac{ P_{17}(P_{20}+5\sqrt2\,P_{19}) }{ P_{18}(P_{19}+5\sqrt2\,P_{18}) }$$

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🔷 Step 4 — Substitute Relations

Using recurrence results:

$$E= \frac{ P_{17}(-10P_{18}) }{ P_{18}(-10P_{17}) }$$

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🔷 Step 5 — Simplify

$$E=1$$

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✅ Final Answer

$$\boxed{1}$$

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⭐ Golden JEE Insight

For sequences involving roots of quadratic equations:

$$P_n+aP_{n-1}+bP_{n-2}=0$$

is the fastest shortcut.

Most difficult-looking expressions collapse after factorisation + recurrence substitution.