Prove the following by the PMI :

n7/ 7 + n5 / 5 + n3 / 3 + n2/ 2 - 37/210n is a positive integer for all n belongs to N.

Question

Prove the following by the PMI :
n7/ 7 + n5 / 5 + n3 / 3 + n2/ 2 - 37/210n is a positive integer
for all n belongs to N

Ajanta Trivedi · Accepted Answer

TPT : $\frac{n^{7}}{7} + \frac{n^{5}}{5} + \frac{n^{3}}{3} + \frac{n^{2}}{2} - \frac{37}{210} n$ is a positive integer for all $n \in N$ .

to prove this by mathematical induction first we will show that it is true for n = 1

and then we will assume that it is true for n = k and by the help of our assumption we will prove that it is true for n = (k+1)

step I;

put n = 1

$\frac{1^{7}}{7} + \frac{1^{5}}{5} + \frac{1^{3}}{3} + \frac{1^{2}}{2} - \frac{37}{210} * 1 = \frac{1}{7} + \frac{1}{5} + \frac{1}{3} + \frac{1}{2} - \frac{37}{210} = \frac{30 + 42 + 70 + 105}{210} - \frac{37}{210} = \frac{247}{210} - \frac{37}{210} = \frac{247 - 37}{210} = \frac{210}{210} = 1$

which is an integer.

step II: let us assume that result holds for n = k

i.e. $\frac{k^{7}}{7} + \frac{k^{5}}{5} + \frac{k^{3}}{3} + \frac{k^{2}}{2} - \frac{37}{210} k = N_{1} (s a y)$ is an integer................(1)

step III:

now we will try prove that it is true for n = k+1

$\frac{(k + 1)^{7}}{7} + \frac{(k + 1)^{5}}{5} + \frac{(k + 1)^{3}}{3} + \frac{(k + 1)^{2}}{2} - \frac{37}{210} (k + 1)$

$\frac{(k + 1)^{7}}{7} = \frac{k^{7}}{7} + (\frac{{}^{7}{C_{1}} k^{6} + ............ + {}^{7}{C_{6}} k}{7}) + \frac{{}^{7}{C_{7}}}{7}$

$∵ {}^{7}{C_{1}}, {}^{7}{C_{2}}, ..........., {}^{7}{C_{6}}$ all are multiple of 7 . therefore

$\frac{{}^{7}{C_{1}} k^{6} + ............ + {}^{7}{C_{6}} k}{7} = N_{2}$ is a positive integer.

thus $\frac{(k + 1)^{7}}{7} = \frac{k^{7}}{7} + N_{2} + \frac{1}{7}$ ...................(2)

$\frac{(k + 1)^{5}}{5} = \frac{k^{5}}{5} + (\frac{{}^{5}{C_{1}} k^{4} + ...... + {}^{5}{C_{4}} k}{5}) + \frac{{}^{5}{C_{5}}}{5} \frac{(k + 1)^{5}}{5} = \frac{k^{5}}{5} + N_{3} + \frac{1}{5}$ ...............(3)

since ${}^{5}{C_{1}}, {}^{5}{C_{2}}, ..........., {}^{5}{C_{4}}$ are multiple of 5. and $N_{3}$ is a positive integer

$\frac{(k + 1)^{3}}{3} = \frac{k^{3}}{3} + (\frac{{}^{3}{C_{1}} k^{2} + {}^{3}{C_{2}} k}{3}) + \frac{{}^{3}{C_{3}}}{3} \frac{(k + 1)^{3}}{3} = \frac{k^{3}}{3} + N_{4} + \frac{1}{3}$ ......................(4) $N_{4}$ is a positive integer

$\frac{(k + 1)^{2}}{2} = \frac{k^{2}}{2} + \frac{2 k}{2} + \frac{1}{2} \frac{(k + 1)^{2}}{2} = \frac{k^{2}}{2} + k + \frac{1}{2} ............. (5) - \frac{37}{210} (k + 1) = - \frac{37}{210} k - \frac{37}{210} ............. (6)$

from eq(2), (3), (4),(5) and (6):

$\frac{(k + 1)^{7}}{7} + \frac{(k + 1)^{5}}{5} + \frac{(k + 1)^{3}}{3} + \frac{(k + 1)^{2}}{2} - \frac{37}{210} (k + 1) = {\frac{k^{7}}{7} + \frac{k^{5}}{5} + \frac{k^{3}}{3} + \frac{k^{2}}{2} - \frac{37}{210} k} + {N_{2} + N_{3} + N_{4} + k} + {\frac{1}{7} + \frac{1}{5} + \frac{1}{3} + \frac{1}{2} - \frac{37}{210}} = N_{1} + {N_{2} + N_{3} + N_{4} + k} + 1$ [from eq(1)]

= sum of positive integers

= a positive integer.

thus we have proved that for n = (k+1); $\frac{n^{7}}{7} + \frac{n^{5}}{5} + \frac{n^{3}}{3} + \frac{n^{2}}{2} - \frac{37}{210} n$ is a positive integer.

thus it is true for all values of n belongs to N.

hope this helps you.

cheers!!

Prove the following by the PMI : n7/ 7 + n5 / 5 + n3 / 3 + n2/ 2 - 37/210n is a positive integer for all n belongs to N.

Prove the following by the PMI :

n7/ 7 + n5 / 5 + n3 / 3 + n2/ 2 - 37/210n is a positive integer for all n belongs to N.