1^2+2^2+3^2+⋯+n^2=(n(n+1)(2n+1))/6
How prove this with defrente way
Answers
Answered by
1
Answer:
Explanation:
using the method of
proof by induction
this involves the following steps
∙
prove true for some value, say n = 1
∙
assume the result is true for n = k
∙
prove true for n = k + 1
n
=
1
→
L
H
S
=
1
2
=
1
and RHS
=
1
6
(
1
+
1
)
(
2
+
1
)
=
1
⇒
result is true for n = 1
assume result is true for n = k
assume
1
2
+
2
2
+
...
.
+
k
2
=
1
6
k
(
k
+
1
)
(
2
k
+
1
)
prove true for n = k + 1
1
2
+
2
2
+
.
.
+
k
2
+
(
k
+
1
)
2
=
1
6
k
(
k
+
1
)
(
2
k
+
1
)
+
(
k
+
1
)
2
=
1
6
(
k
+
1
)
[
k
(
2
k
+
1
)
+
6
(
k
+
1
)
]
=
1
6
(
k
+
1
)
(
2
k
2
+
7
k
+
6
)
=
1
6
(
k
+
1
)
(
k
+
2
)
(
2
k
+
3
)
=
1
6
n
(
n
+
1
)
(
2
n
+
1
)
→
with
n
=
k
+
1
⇒
result is true for n = k + 1
⇒1+2+3+....+n2=16n(n+1)(2n+1)
Answered by
0
Step-by-step explanation:
(k-1)^3=k^3-3k^2+3k-1
k^3-(k-1)^3=3k^2-3k+1
on adding all terms
2^3-1^3 =3×1^2 -3×1 +1
3^3-2^3 =3×2^2 -3×2+1
.
.
.
some terms will cut out and
3×sum of n^2=n^3+3(n)(n+1)/2-n
sum of n^2=n(n+1)(2n+1)/6
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