Question

Observe the following pattern

12 = 1/6 (1×(1+1)×(2×1+1))

12+22 = 1/6 (2×(2+1)×(2×2+1)))

12+22+32 = 1/6 (3×(3+1)×(2×3+1)))

12+22+32+42 = 1/6 (4×(4+1)×(2×4+1)))

And find the values of each of the following:

(i) 12+22+32+42+…+102

(ii) 52+62+72+82+92+102+112+122

Answer 1

RHS = *1/6 [(No. of terms in L.H.) × (No. of terms + 1) × (2 × No. of terms + 1)]**

(i) 12+22+32+42+…+102

= 1/6 (10×(10+1)×(2×10+1))

= 1/6 (2310)

= 385

(ii) 52+62+72+82+92+102+112+122

= 12+22+32+…+122 – (12+22+32+42)

= 1/6 (12×(12+1)×(2×12+1)) – 1/6 (4×(4+1)×(2×4+1))

= 650-30

= 620