Let \[P\text{ }\left( n \right):\text{ }{{7}^{2n}}~+\text{ }{{2}^{3n\text{ }\text{ }3}}.\text{ }3n\text{ }\text{ }1\] is divisible by 25 Let us check for \[n\text{ }=\text{ }1,\] \[P\text{ }\left( 1...
Solve : n (n + 1) (n + 5) is a multiple of 3 for all n ϵ N.
Let, \[P\text{ }\left( n \right):\text{ }n\text{ }\left( n\text{ }+\text{ }1 \right)\text{ }\left( n\text{ }+\text{ }5 \right)\] is a multiple of 3 Let us check for \[n\text{ }=\text{ }1,\]...
Solve: (ab) n = an bn for all n ϵ N
Let , \[P\text{ }\left( n \right):\text{ }{{\left( ab \right)}^{~n}}~=\text{ }{{a}^{n}}~{{b}^{n}}\] Let us check for \[n\text{ }=\text{ }1,\] \[P\text{ }\left( 1 \right):\text{ }{{\left( ab...
prove: 32n + 2 – 8n – 9 is divisible by 8 for all n ϵ N
\[Let\text{ }P\text{ }\left( n \right):\text{ }{{3}^{2n\text{ }+\text{ }2}}~\text{ }8n\text{ }\text{ }9\text{ }is\text{ }divisible\text{ }by\text{ }8\] Let us check for \[n\text{ }=\text{ }1,\]...
Solve : 52n + 2 – 24n – 25 is divisible by 576 for all n ϵ N
Let , \[P\text{ }\left( n \right):\text{ }{{5}^{2n\text{ }+\text{ }2}}~\text{ }24n\text{ }\text{ }25\text{ }is\text{ }divisible\text{ }by\text{ }576\] Let us check for \[n\text{ }=\text{ }1,\]...
Solve:32n + 7 is divisible by 8 for all n ϵ N
Let, \[~P\text{ }\left( n \right):\text{ }{{3}^{2n}}~+\text{ }7\text{ }is\text{ }divisible\text{ }by\text{ }8\] Let us check for \[n\text{ }=\text{ }1,\] \[P\text{ }\left( 1 \right):\text{...
Solve: 52n – 1 is divisible by 24 for all n ϵ N
Let , \[P\text{ }\left( n \right):~{{5}^{2n}}~\text{ }1\text{ }is\text{ }divisible\text{ }by\text{ }24\] Let us check for \[n\text{ }=\text{ }1,\] \[P\text{ }\left( 1 \right):\text{...
Solve: a + (a + d) + (a + 2d) + … + (a + (n-1)d) = n/2 [2a + (n-1)d]
Let , \[P\text{ }\left( n \right):~a\text{ }+\text{ }\left( a\text{ }+\text{ }d \right)\text{ }+\text{ }\left( a\text{ }+\text{ }2d \right)\text{ }+\text{ }\ldots \text{ }+\text{ }\left( a\text{...
Solve: a + ar + ar2 + … + arn – 1 = a [(rn – 1)/(r – 1)], r ≠ 1
Let , \[P\text{ }\left( n \right):~a\text{ }+\text{ }ar\text{ }+\text{ }a{{r}^{2}}~+\text{ }\ldots \text{ }+\text{ }a{{r}^{n\text{ }\text{ }1}}~=\text{ }a\text{ }\left[ \left( {{r}^{n}}~\text{ }1...
Solve:1/2 + 1/4 + 1/8 + … + 1/2n = 1 – 1/2n
Let , \[P\text{ }\left( n \right):~1/2\text{ }+\text{ }1/4\text{ }+\text{ }1/8\text{ }+\text{ }\ldots \text{ }+\text{ }1/{{2}^{n}}~=\text{ }1\text{ }\text{ }1/{{2}^{n}}\] Let us check for \[n\text{...
Solve: 1.2 + 2.3 + 3.4 + … + n(n+1) = [n (n+1) (n+2)] / 3
Let , P(n): given equation Let us check for \[n\text{ }=\text{ }1,\] \[P\text{ }\left( 1 \right):\text{ }1\left( 1+1 \right)\text{ }=\text{ }\left[ 1\left( 1+1 \right)\text{ }\left( 1+2 \right)...
Prove:1.3 + 3.5 + 5.7 + … + (2n – 1) (2n + 1) = n(4n2 + 6n – 1)/3
Let, P(n) be the given equation. Let us check for \[n\text{ }=\text{ }1,\] \[P\text{ }\left( 1 \right):\text{ }\left( 2.1\text{ }\text{ }1 \right)\text{ }\left( 2.1\text{ }+\text{ }1 \right)\text{...
Prove:1.3 + 2.4 + 3.5 + … + n. (n+2) = 1/6 n (n+1) (2n+7)
Let, \[~P\text{ }\left( n \right):\text{ }1.3\text{ }+\text{ }2.4\text{ }+\text{ }3.5\text{ }+\text{ }\ldots \text{ }+\text{ }n.\text{ }\left( n+2 \right)\text{ }=\text{ }1/6\text{ }n\text{ }\left(...
Prove:2 + 5 + 8 + 11 + … + (3n – 1) = 1/2 n (3n + 1)
Let, \[P\text{ }\left( n \right)\text{ }=\text{ }2\text{ }+\text{ }5\text{ }+\text{ }8\text{ }+\text{ }11\text{ }+\text{ }\ldots \text{ }+\text{ }\left( 3n\text{ }\text{ }1 \right)\text{ }=\text{...
Prove: 1.2 + 2.22 + 3.23 + … + n.2n = (n–1) 2n + 1 + 2
\[~P\text{ }\left( n \right)\text{ }=~1.2\text{ }+\text{ }{{2.2}^{2}}~+\text{ }{{3.2}^{3}}~+\text{ }\ldots \text{ }+\text{ }n{{.2}^{n~}}=\text{ }\left( n1 \right)\text{ }{{2}^{n\text{ }+\text{...
Prove:1/3.7 + 1/7.11 + 1/11.15 + … + 1/(4n-1)(4n+3) = n/3(4n+3)
Let. \[P\text{ }\left( n \right)\text{ }=\text{ }1/3.7\text{ }+\text{ }1/7.11\text{ }+\text{ }1/11.15\text{ }+\text{ }\ldots \text{ }+\text{ }1/\left( 4n-1 \right)\left( 4n+3 \right)\text{ }=\text{...
Prove: 1/3.5 + 1/5.7 + 1/7.9 + … + 1/(2n+1)(2n+3) = n/3(2n+3)
Let, \[P\text{ }\left( n \right)\text{ }=\text{ }1/3.5\text{ } +\text{ }1/5.7\text{ }+\text{ }1/7.9\text{ }+\text{ }\ldots \text{ } +\text{ }1/\left( 2n+1 \right)\left( 2n+3 \right)\text{ }=\text{...
Prove: 1/1.4 + 1/4.7 + 1/7.10 + … + 1/(3n-2)(3n+1) = n/3n+1
Let, $~P\text{ }\left( n \right)\text{ }=\text{ }1/1.4\text{ }+\text{ }1/4.7\text{ }+\text{ }1/7.10\text{ }+\text{ }\ldots \text{ }+\text{ }1/\left( 3n-2 \right)\left( 3n+1 \right)\text{ }=\text{...
Prove:1/2.5 + 1/5.8 + 1/8.11 + … + 1/(3n-1) (3n+2) = n/(6n+4)
Let , \[P\text{ }\left( n \right)\text{ }=\text{ }1/2.5\text{ }+\text{ }1/5.8\text{ }+\text{ }1/8.11\text{ }+\text{ }\ldots \text{ }+\text{ }1/\left( 3n-1 \right)\text{ }\left( 3n+2 \right)\text{...
prove: 1 + 3 + 5 + … + (2n – 1) = n2 i.e., the sum of first n odd natural numbers is n2.
Let, \[~P\text{ }\left( n \right):\text{ }1\text{ }+\text{ }3\text{ }+\text{ }5\text{ }+\text{ }\ldots \text{ }+\text{ }\left( 2n\text{ }\text{ }1 \right)\text{ }=\text{ }{{n}^{2}}\] Let us check P...
Prove: 1/1.2 + 1/2.3 + 1/3.4 + … + 1/n(n+1) = n/(n+1)
Let, \[P\text{ }\left( n \right)\text{ }=\text{ }1/1.2\text{ }+\text{ }1/2.3\text{ }+\text{ }1/3.4\text{ }+\text{ }\ldots \text{ }+\text{ }1/n\left( n+1 \right)\text{ }=\text{ }n/\left( n+1...
Prove:1 + 3 + 32 + … + 3n-1 = (3n – 1)/2
Let, \[~P\text{ }\left( n \right)\text{ }=\text{ }1\text{ }+\text{ }3\text{ }+\text{ }{{3}^{2}}~+\text{ }\text{ }\text{ }\text{ }\text{ }+\text{ }{{3}^{n\text{ }\text{ }1}}~=\text{ }\left(...
Prove:12 + 22 + 32 + … + n2 = [n (n+1) (2n+1)]/6
Let, \[P\text{ }\left( n \right)\text{ }=\text{ }{{1}^{2}}~+\text{ }{{2}^{2}}~+\text{ }{{3}^{2}}~+\text{ }\ldots \text{ }+\text{ }{{n}^{2}}~=\text{ }\left[ n\text{ }\left( n+1 \right)\text{ }\left(...
1 + 2 + 3 + … + n = n (n +1)/2 i.e., the sum of the first n natural numbers is n (n + 1)/2.
Let, \[P\text{ }\left( n \right)\text{ }=\text{ }1\text{ }+\text{ }2\text{ }+\text{ }3\text{ }+\text{ }\ldots ..\text{ }+\text{ }n\text{ }=\text{ }n\text{ }\left( n\text{ }+1 \right)/2\] For,...
Find x in each of the following: (iii) 1/6! + 1/7! = x/8!
Solution: We can write that: 8! = 8 × 7 × 6! 7! = 7 × 6! So by making use of these values, $ 1/6!\text{ }+\text{ }1/7!\text{ }=\text{ }x/8! $ $ 1/6!\text{ }+\text{ }1/\left( 7\times 6! \right)\text{...
If P (n) is the statement “n2 – n + 41 is prime”, prove that P (1), P (2) and P (3) are true. Prove also that P (41) is not true.
According to ques: \[P\left( n \right)\text{ }=\text{ }{{n}^{2}}~\text{ }n\text{ }+\text{ }41\] is prime. \[P\left( n \right)\text{ }=\text{ }{{n}^{2}}~\text{ }n\text{ }+\text{ }41\] Or,...
Give an example of a statement P (n) such that it is true for all n ϵ N
Let , \[P\text{ }\left( n \right)\text{ }=\text{ }1\text{ }+\text{ }2\text{ }+\text{ }3\text{ }+\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }+\text{ }n\text{ }=\text{ }n\left( n+1 \right)/2~\]...
If P (n) is the statement “n2 + n” is even”, and if P (r) is true, then P (r + 1) is true
According to ques: \[P\text{ }\left( n \right)\text{ }=\text{ }{{n}^{2}}~+\text{ }n\] is even and P (r) is true, then \[{{r}^{2}}~+\text{ }r\] is even Let \[{{r}^{2}}~+\text{ }r\text{ }=\text{...
If P (n) is the statement “2n ≥ 3n”, and if P (r) is true, prove that P (r + 1) is true.
According to ques: \[P\text{ }\left( n \right)\text{ }=\text{ }{{2}^{n}}~\ge \text{ }3n\] and p(r) is true. Now, \[P\text{ }\left( n \right)\text{ }=\text{ }{{2}^{n}}~\ge \text{ }3n\] Since,...
If P (n) is the statement “n3 + n is divisible by 3”, prove that P (3) is true but P (4) is not true.
According to ques: \[P\text{ }\left( n \right)\text{ }=\text{ }{{n}^{3}}~+\text{ }n\] is divisible by 3 Now, \[P\text{ }\left( n \right)\text{ }=\text{ }{{n}^{3}}~+\text{ }n\] Hence,...
If P (n) is the statement “n (n + 1) is even”, then what is P (3)?
According to ques: \[P\text{ }\left( n \right)\text{ }=\text{ }n\text{ }\left( n\text{ }+\text{ }1 \right)\] is even. Hence, \[P\text{ }\left( 3 \right)\text{ }=\text{ }3\text{ }\left( 3\text{...