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Worked solution

Evaluate three finite sums

Evaluate: (i) 12+22+32+42+52+62+72+82+92+102+112+122+132+142+1521^2+2^2+3^2+4^2+5^2+6^2+7^2+8^2+9^2+10^2+11^2+12^2+13^2+14^2+15^2; (ii) ∑r=120(3−2r)\sum_{r=1}^{20}(3-2r); (iii) ∑r=0152r\sum_{r=0}^{15}2^r.

CJC · Sequences and Series · Self-Practice 5

(i) Fifteen squares

Answer

12401240.

Add all fifteen squared terms

(1+4+9+16+25)+(36+49+64+81+100)+(121+144+169+196+225)=55+330+855=1240\begin{aligned}&(1+4+9+16+25)\\ &{}+(36+49+64+81+100)\\ &{}+(121+144+169+196+225)\\ & =55+330+855\\ & =1240\end{aligned}

The source prints the answer only. This complete grouping is authored working.

How did this part go?

(ii) Twenty odd terms

Answer

−360-360.

Substitute inclusive bounds and pair terms

∑r=120(3−2r)=1+(−1)+(−3)+⋯+(−37)=10(1−37)=−360\begin{aligned}&\sum_{r=1}^{20}(3-2r)\\ & =1+(-1)+(-3)+\cdots+(-37)\\ & =10(1-37)\\ & =-360\end{aligned}

Authored working: the twenty terms run from 1 to −37; opposite-end pairs each sum to −36.

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(iii) Sixteen powers of two

Answer

6553565535.

Keep the zeroth-power term

T=1+2+22+⋯+215,2T−T=216−1,T=65536−1=65535.\begin{aligned}T&=1+2+2^2+\cdots+2^{15},\\2T-T&=2^{16}-1,\\T&=65536-1=65535.\end{aligned}

Authored working: subtract the sum from twice itself; shared powers cancel, leaving the new final power and the initial 1.

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Solution notes

The three numerical answers are printed; these derivations supplement the source answer key.