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Worked solution

Write fourth powers and alternating integers as sums

Express in sigma notation: (i) 14+24+⋯+n4+(n+1)41^4+2^4+\cdots+n^4+(n+1)^4; (ii) −1+2−3+4−5+6-1+2-3+4-5+6.

CJC · Sequences and Series · Self-Practice 4

(i) Fourth powers

Answer

∑r=1n+1r4\sum_{r=1}^{n+1}r^4 or ∑r=0n(r+1)4\sum_{r=0}^{n}(r+1)^4.

Keep the last power and the number of terms

r=1,…,n+1:r4=14,…,(n+1)4,r=0,…,n:(r+1)4=14,…,(n+1)4.\begin{aligned}r=1,\ldots,n+1&:\quad r^4=1^4,\ldots,(n+1)^4,\\r=0,\ldots,n&:\quad(r+1)^4=1^4,\ldots,(n+1)^4.\end{aligned}

Both forms are printed. The second shifts the counter down by one while keeping the same terms.

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(ii) Alternating integers

Answer

∑r=16(−1)rr\sum_{r=1}^{6}(-1)^r r or ∑r=27(−1)r−1(r−1)\sum_{r=2}^{7}(-1)^{r-1}(r-1).

Check both signs at the endpoints

(−1)1(1)=−1,(−1)6(6)=6,(−1)2−1(2−1)=−1,(−1)7−1(7−1)=6(-1)^1(1)=-1,\quad(-1)^6(6)=6,\qquad(-1)^{2-1}(2-1)=-1,\quad(-1)^{7-1}(7-1)=6

The first formula gives −1 at r=1 and +6 at r=6. The second gives the same endpoints at r=2 and r=7.

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