Question

Simplify...... Hey guys i am Ichha Raina of class 9 ....i wanna get my question solved . In the attachment it is Q.22... Plz pay attention I ONLU NEED THE ANSWER OF Q.22.♡

Answer 1

$\huge\mathtt\red{Solution}$

In+2.(n−1)+3(n−2)+......n.1

⇒tr=r(n−r)

∴    ∑r=1ntr=∑r=1nr(n−r)

=∑r=1n(nr−r2)

=n.2n(n+1)−6n(n+1)(2n+1)

=2n2(n+1)−6(n2+n)(2n+1)=2n3+n−(62n3+3n2+n)=6n3−3n2+2n

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Answer 2

$\huge\bigstar\underline\mathfrak{Answer:-}$

$\longrightarrow$In+2.(n−1)+3(n−2)+......n.1

⇒tr=r(n−r)

∴ ∑r=1ntr=∑r=1nr(n−r)

$\longrightarrow$∑r=1n(nr−r2)

$\longrightarrow$n.2n(n+1)−6n(n+1)(2n+1)

$\longrightarrow$2n2(n+1)−6(n2+n)(2n+1)=2n3+n−(62n3+3n2+n)=6n3−3n2+2n