Gujarat BoardEnglish MediumSTD 11 ScienceMATHSBINOMIAL THEOREM1 Mark
Question
Using binomial theorem, prove that $6^n–5n$ always leaves remainder $1$ when divided by $25$.
✓
Answer
For two numbers a and b if we can find numbers q and r such that $a = bq + r$, then we say that b divides a with q as quotient and r as remainder. Therefore, in order to show that $n^6$– 5n leaves remainder 1 when divided by $25$, now we have to prove that $n^6– 5n = 25k + 1$, where k is some natural number.
Now,we have
$(1+a)^n={ }^n C_0+{ }^n C_1 a+{ }^n C_2 a^2+\ldots+{ }^n C_n a^n$
For $a=5$, we obtain
$(1+5)^n={ }^n C_0+{ }^n C_1 5+{ }^n C_2 5^2+\ldots+{ }^n C_n 5^n$
i.e. $(6)^n=1+5 n+5^2 \cdot{ }^n C_2+5^3 \cdot{ }^n C_3+\ldots+5^n$
i.e. $n 6^n-5 n=1+5^2\left({ }^n C_2+{ }^n C_3 5+\ldots+5^{n-2}\right)$
or $6^n-5 n=1+25\left({ }^n C_2+5 .{ }^n C_3+\ldots+5^{n-2}\right)$
or $6^n-5 n=25 k+1$ where $k={ }^n C_2+5 .{ }^n C_3+\ldots+5^{n-2}$.
Therefore, This shows that when divided by $25,6^n-5 n$ leaves remainder 1 .
Therefore,This shows that when divided by $25, 6^n – 5n$ leaves remainder 1.
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