Question

Show that
\[\sum_{n+1}^{2n}(6r+4)=n(9n+7)\]

Answer 1

hint\[\sum_{r=n+1}^{2n} (6r+4)=\sum_{r=1}^{2n} (6r+4)-\sum_{r=1}^{n} (6r+4)\]
and use this\[\sum_{k=1}^{n} k=\frac{n(n+1)}{2} \]

Answer 2

Arithmetic Sequence with r=6
The first term is 4 + 6 (1 + n)
the last term is 4 + 6 (1 + n) + 6 (n - 1)

\[
S =\frac n 2 ( a_1 + a_n)=
\\
S =\frac n 2 (4 + 6 (1 + n) +4 + 6 (1 + n) + 6 (n - 1))=n (7 + 9 n)

\]

Answer 3

lol...much better and simpler...thank u professor

Answer 4

yw

Answer 5

@mukushla you do you subtract??