CBSE 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
$(1 + a)^n = ^nC_0 + ^nC_1a + ^nC_2a^2 + ... + ^nC_na^n$
For a = 5, we obtain
$(1 + 5)^n = ^nC_0 + ^nC_15 + ^nC_25^2 + ... + ^nC_n5^n$
i.e. $(6)^n = 1 + 5n + 5^2.^nC_2 + 5^3.^nC_3 + ... + 5^n$
i.e. $ n 6^n – 5n = 1 + 5^2 (^nC_2 + ^nC_35 + ... + 5^{n-2})$
or $6^n – 5n = 1 + 25 (^nC_2 + 5.^nC_3 + ... + 5^{n-2})$
or $6^n – 5n = 25k+1$ where $k = ^nC_2 + 5.^nC_3 + ... + 5^{n–2}.$
Therefore,This shows that when divided by $25, 6^n – 5n$ leaves remainder $1.$
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