Question

The number ${51^{49}} + {51^{48}} + {51^{47}} + ....... + 51 + 1$ is divisible by
(A) $10$
(B) $20$
(C) $25$
(D) $50$

Answer 1

Hint:In the given question, we are required to find out that the summation of the series given to us is divisible by which of the number in the four options given to us. The series given to us is ${51^{49}} + {51^{48}} + {51^{47}} + ....... + 51 + 1$ . We have to first find the value of summation of the given series and then find it if it is divisible by any of the given numbers.

Complete step by step answer:
The given number in the form of series is ${51^{49}} + {51^{48}} + {51^{47}} + ....... + 51 + 1$ .
So, we can observe that the given series is in a geometric progression.Here, we can observe that the series is in a geometric progression (GP) whose first term is $1$ and common ratio is $51$ and total number of terms is $50$.So, using the sum of n terms formula for a geometric progression, we get,
${S_n} = a\dfrac{{\left( {{r^n} - 1} \right)}}{{\left( {r - 1} \right)}}$
$ \Rightarrow {S_{50}} = \left( 1 \right)\dfrac{{\left( {{{51}^{50}} - 1} \right)}}{{50}}$
Now, we need to find out that the expression ${51^{49}} + {51^{48}} + {51^{47}} + ....... + 51 + 1$ is divisibly]e by which number. We know that $51 = 50 + 1$. Hence, \[{51^{50}} = {\left( {50 + 1} \right)^{50}}\].

Now, using binomial theorem, we get,
${51^{50}}{ = ^{50}}{C_0}{\left( {50} \right)^{50}}{ + ^{50}}{C_1}{\left( {50} \right)^{49}}....{ + ^{50}}{C_{49}}\left( {50} \right){ + ^{50}}{C_{50}}$
Hence, ${S_{50}} = \left( 1 \right)\dfrac{{\left( {^{50}{C_0}{{\left( {50} \right)}^{50}}{ + ^{50}}{C_1}{{\left( {50} \right)}^{49}}....{ + ^{50}}{C_{49}}\left( {50} \right){ + ^{50}}{C_{50}} - 1} \right)}}{{50}}$
Since, $^{50}{C_{50}} = 1$. Hence, we can cancel $^{50}{C_{50}}$ with $ - 1$. So, we get,
$ \Rightarrow {S_{50}} = \dfrac{{\left( {^{50}{C_0}{{\left( {50} \right)}^{50}}{ + ^{50}}{C_1}{{\left( {50} \right)}^{49}}....{ + ^{50}}{C_{49}}\left( {50} \right)} \right)}}{{50}}$
$ \Rightarrow {S_{50}}{ = ^{50}}{C_0}{\left( {50} \right)^{49}}{ + ^{50}}{C_1}{\left( {50} \right)^{48}}....{ + ^{50}}{C_{49}}$

Now, we know that the value of $^{50}{C_{49}}$ is $50$.
$ \Rightarrow {S_{50}}{ = ^{50}}{C_0}{\left( {50} \right)^{49}}{ + ^{50}}{C_1}{\left( {50} \right)^{48}}.... + 50$
Now, all the terms of the above expression are divisible by $50$ individually. So, their sum will also be divisible by $50$.
$ \Rightarrow {S_{50}} = 50n$ where n is any integer.
So, the summation of the given series is divisible by $50$. Since, the sum of series is divisible by $50$, so it is also divisible by $10$ and $25$.

Now, we have to check whether the given summation is also divisible by $20$ or not.
So, ${S_{50}}{ = ^{50}}{C_0}{\left( {50} \right)^{49}}{ + ^{50}}{C_1}{\left( {50} \right)^{48}}.... + 50$
We take out $50$ common from the sum, we get,
$ \Rightarrow {S_{50}} = 50\left[ {^{50}{C_0}{{\left( {50} \right)}^{48}}{ + ^{50}}{C_1}{{\left( {50} \right)}^{47}}.... + 1} \right]$
Now, we have to check whether the whole bracket in the above expression is divisible by $2$or not because if it is divisible by $2$, then the whole summation would be divisible by $100$ as well and hence by $20$ also.On observing the terms inside the bracket, we see that all the terms are even except the last term which is one. Hence, the sum is not divisible by $20$.

Note: Here we see that the given number is in the form of a series which is divisible by many numbers but we have to answer according to the numbers given in the options as this is an objective type question. Question involves the concepts of binomial theorem and geometric progressions.