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Thanks guest, here is a little more insight for others to see what you have done.

 

(a) Compute the sum 101^2 - 97^2 + 93^2 - 89^2 + .... + 5^2 - 1^2

 

I paired them first

(101^2 - 97^2) + (93^2 - 89^2) + .... + (5^2 - 1^2)

 

then turned them around

(5^2 - 1^2)+ ..................   +  (93^2 - 89^2) +  (101^2 - 97^2) 

 

\(T_1=(5^2 - 1^2) = [(3+2)^2 - (3-2)^2]\\ T_2=(13^2-9^2)= [(11+2)^2 - (11-2)^2]\\ ...\\ T_n= [((8n-5)+2)^2 - ((8n-5)-2)^2]\\ ...\\ T_{13}=(101^2-97^2)\\\quad= [(8*13-5+2)^2 - (8*13-5-2)^2]\\ \quad= [(99+2)^2 - (99-2)^2]\quad \text{just checking... it is good}\\ \)

 

\(T_n= [((8n-5)+2)^2 - ((8n-5)-2)^2]\\ T_n= [(8n-3)^2 - (8n-7)^2]\\ T_n= [(8n)^2-48n+9)] - [(8n)^2-112n+49)]\\ T_n= [-48n+9)] - [-112n+49)]\\ T_n= -48n+9 +112n-49\\ T_n= 64n-40\\ \text{This is an AP}\\ a=24, d=64,\;n=13\\ S_n=\frac{n}{2}(2a+(n-1)d]\\ S_{13}=\frac{13}{2}(2*24+12*64]\\ S_{13}=13(24+6*64]\\ S_{13}=13*408\\ S_{13}=5304\\\)

 

 

 

 

LaTex:

T_1=(5^2 - 1^2) = [(3+2)^2 - (3-2)^2]\\
T_2=(13^2-9^2)=  [(11+2)^2 - (11-2)^2]\\
...\\
T_n= [((8n-5)+2)^2 - ((8n-5)-2)^2]\\
...\\
T_{13}=(101^2-97^2)\\\quad=  [(8*13-5+2)^2 - (8*13-5-2)^2]\\
\quad=  [(99+2)^2 - (99-2)^2]\quad \text{just checking... it is good}\\

 

T_n= [((8n-5)+2)^2 - ((8n-5)-2)^2]\\
T_n= [(8n-3)^2 - (8n-7)^2]\\
T_n= [(8n)^2-48n+9)] - [(8n)^2-112n+49)]\\
T_n= [-48n+9)] - [-112n+49)]\\
T_n= -48n+9 +112n-49\\
T_n= 64n-40\\
\text{This is an AP}\\
a=24, d=64,\;n=13\\
S_n=\frac{n}{2}(2a+(n-1)d]\\
S_{13}=\frac{13}{2}(2*24+12*64]\\
S_{13}=13(24+6*64]\\
S_{13}=13*408\\
S_{13}=5304\\

Aug 27, 2020

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