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Find
∑
k
=
2
7
k
2
Find
∑
k
=
2
7
k
3
Hence
, find
∑
k
=
2
7
k
2
+
k
3
(
2
2
+
2
3
)
+
(
3
2
+
3
3
)
+
…
+
(
7
2
+
7
3
)
=
(
2
2
+
3
2
+
…
+
7
2
)
+
(
2
3
+
3
3
+
…
+
7
3
)