1+2+3+...+n=1/2n (n+1)
Answers
Answer:
to prove by induction
#1+2+3+..n=1/2n(n+1)#
#color(red)((1) " verify for " n=1)#
#LHS=1#
#RHS=1/2xx1xx(1+1)=1/2xx1xx2=1#
#:. "true for "n=1#
#color(red)((2)" to prove "T_k=>T_(k+1))#
#"assume true for "T_k=1/2k(k+1)#
to prove #T_(k+1)=1/2(k+1)(k+2)#
add the next term
#RHS=1/2k(k+1)+(k+1)#
#=(k+1)(1/2k+1)#
#=1/2(k+1)(k+2)=T_(k+1)" as required"#
#:. T_k=>T_(k+1)#
#color(red)((3) " conclusion"#
#(i) " "T_1" is true"#
#(ii)" " T_k=>T_(k+1)#
#:. T_1=>T_2#
#T_2=>T_3" etc."#
therefore by induction true for values
#1,2,3,...#
1
+
2
+
3
+
.
.
n
=
1
2
n
(
n
+
1
)
(
1
)
verify for
n
=
1
L
H
S
=
1
R
H
S
=
1
2
×
1
×
(
1
+
1
)
=
1
2
×
1
×
2
=
1
∴
true for
n
=
1
(
2
)
to prove
T
k
⇒
T
k
+
1
assume true for
T
k
=
1
2
k
(
k
+
1
)
to prove
T
k
+
1
=
1
2
(
k
+
1
)
(
k
+
2
)
add the next term
R
H
S
=
1
2
k
(
k
+
1
)
+
(
k
+
1
)
=
(
k
+
1
)
(
1
2
k
+
1
)
=
1
2
(
k
+
1
)
(
k
+
2
)
=
T
k
+
1
as required
∴
T
k
⇒
T
k
+
1
(
3
)
conclusion
(
i
)
T
1
is true
(
i
i
)
T
k
⇒
T
k
+
1
∴
T
1
⇒
T
2
T
2
⇒
T
3
etc.
therefore by induction true for values
1
,
2
,
3
,
...
and so
∀
n
∈
N
Step-by-step explanation:
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