Prove that - r-1 n r- n n-1 - 2

A] Let ∑_r=1^nr = T(n) T(n)+T(n) = 1+2+....+ (n 1)+n+n+ ....+1 2T(n)= (1+n) + (2+n 1) + .....+(n+1) 2T(n)= (n+1)+(n+1) +....+(n+1) 2T(n ...

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Byju's Answer

Standard VII

Mathematics

(a-b)^2 Expansion and Visualisation

Prove that ...

Question

Prove that $n \sum r = 1 r = \frac{n (n + 1)}{2}$

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Solution

A] Let $\sum_{r = 1}^{n} r = T (n)$
$T (n) + T (n) = 1 + 2 + . . . . + (n - 1) + n + n + . . . . + 1$
$2 T (n) = (1 + n) + (2 + n - 1) + . . . . . + (n + 1)$
$2 T (n) = (n + 1) + (n + 1) + . . . . + (n + 1)$
$2 T (n) = n (n + 1)$
$T (n) = \frac{n (n + 1)}{2}$
$\sum_{i = 1}^{n} r = \frac{n (n + 1)}{2}$
Hence proved

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Q. Prove that:
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Prove that n∑r=1r=n(n+1)2

A] Let ∑nr=1r=T(n)T(n)+T(n)=1+2+....+(n−1)+n+n+....+12T(n)=(1+n)+(2+n−1)+.....+(n+1)2T(n)=(n+1)+(n+1)+....+(n+1)2T(n)=n(n+1)T(n)=n(n+1)2∑ni=1r=n(n+1)2Hence proved

Prove that $n \sum r = 1 r = \frac{n (n + 1)}{2}$