Το show 1² + 2² + 3² +4² +5² + To proove by induction; + n² n(n+1)(2n+1) 6

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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1² + 2² + 3² +
MEK
By Induction,
2
+ K² + (k+ 11²
EIN
1² +2²+
=
+ n² =
=
=
so statement is true
= (K+1) { 2K²+ 7k+6}
6
-
(K+1) { K(2K+1) + 6(k+1}}
6
=
(K+1) { 2K²+k+ 6K+6}
6
=
· (K+¹) { 2K² +3K+ 4K+6}
G
(k+1) { (2k + 3) (k+2)}
6
(k+1) (K+1+1) (2(K+1)+1)
6
for
n = (k+1)
n(n+1) (2n+1)
6
when
Hence prooved.
true for
Transcribed Image Text:1² + 2² + 3² + MEK By Induction, 2 + K² + (k+ 11² EIN 1² +2²+ = + n² = = = so statement is true = (K+1) { 2K²+ 7k+6} 6 - (K+1) { K(2K+1) + 6(k+1}} 6 = (K+1) { 2K²+k+ 6K+6} 6 = · (K+¹) { 2K² +3K+ 4K+6} G (k+1) { (2k + 3) (k+2)} 6 (k+1) (K+1+1) (2(K+1)+1) 6 for n = (k+1) n(n+1) (2n+1) 6 when Hence prooved. true for
To show
To proove
1² + 2² + 3² +4² +5².
by
Base Case : 1
Let
>>
2
1²+ 2² + 3² +
n = 1
So statement
inductive case :
Now
Now n(n+1)(2n+1)
6
Let statement
we will
n=k+1
induction;
when
true for
statement is true for
1² + 2² + 3² +
show
is true
true
1²+ 2² + 3² +
n².
=
=
=
that
=
far
for
n(n+1)(2+1)
6
1² =
n = 1
1(1+1)(2+1)
6
+ n²
2
+k²
MEK
n = K
state ment
n≤k
2
+ k² + (k+1) ²
k(k+1)(2k+1)
6
(x+1) {
=
K(K+1)(2K+1)
6
is true for
+ (k+1)²
K(2K+1) + (k+1) }
By
0
all
Transcribed Image Text:To show To proove 1² + 2² + 3² +4² +5². by Base Case : 1 Let >> 2 1²+ 2² + 3² + n = 1 So statement inductive case : Now Now n(n+1)(2n+1) 6 Let statement we will n=k+1 induction; when true for statement is true for 1² + 2² + 3² + show is true true 1²+ 2² + 3² + n². = = = that = far for n(n+1)(2+1) 6 1² = n = 1 1(1+1)(2+1) 6 + n² 2 +k² MEK n = K state ment n≤k 2 + k² + (k+1) ² k(k+1)(2k+1) 6 (x+1) { = K(K+1)(2K+1) 6 is true for + (k+1)² K(2K+1) + (k+1) } By 0 all
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